8

Q1. If A, B, C are interior angles of ABC, show that: cosec² - tan² = 1

Solution

= 1 (since, sec² = 1 + tan²)

Q2. If acos θ + bsin θ = 4 and asin θ - bcos θ = 3, then a² + b² is

1) 7
2) 25
3) 12
4) None

Solution

acosθ plus θsinθ equals 4 squaring space both space sides comma open parentheses acosθ plus bsinθ close parentheses squared equals 4 squared rightwards double arrow straight a squared cos squared straight theta plus straight b squared sin squared straight theta plus 2 abcosθsinθ equals 16... left parenthesis straight i right parenthesis asinθ minus bcosθ equals 3 Squaring comma open parentheses asinθ minus bcosθ close parentheses squared equals 3 squared rightwards double arrow straight a squared cos squared straight theta plus straight b squared sin squared straight theta minus 2 abcosθsinθ equals 9.... left parenthesis ii right parenthesis Adding space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis rightwards double arrow straight a squared open parentheses cos squared straight theta plus sin squared straight theta close parentheses plus straight b squared open parentheses cos squared straight theta plus sin squared straight theta close parentheses equals 25 rightwards double arrow open parentheses straight a squared plus straight b squared close parentheses open parentheses cos squared straight theta plus sin squared straight theta close parentheses equals 25 rightwards double arrow straight a squared plus straight b squared equals 25

Q3. If

cot space straight theta equals 7 over 8

, then

equals to:

Solution

Q4. If tan = 3 sin, find the value of sin² - cos²

Solution

Q5. The express sin A in terms of cot A is:

Solution

Q6. Prove that (cosec - cot)² =

Solution

Q7. If + β=90⁰ and :β=2:1, then sin: sin β=

1) 1:
2) 1:3
3) 1:2
4) :1

Solution

Q8. If 3 cos = 1, then the value of cosec is :

Solution

cos _{= $1 third equals base over hypotenuse$} Let base = k and hypotenuse = 3k Perpendicular =

Q9.

If figure, ABC is right angled triangle, right angled at C. D is mid-point of BC. Show that

Solution

In ABC,

In ACD,

begin mathsize 12px style tan space straight ϕ equals AC over CD end style

Q10. Without using trigonometric table prove that: tan 1^o tan 11^o tan 21^o tan 69^o tan 79^o tan 89^o = 1

Solution

tan 1^o tan 11^o tan 21^o tan 69^o tan 79^o tan 89^o = tan (90^o - 89^o) tan (90^o - 79^o) tan (90^o - 69^o) tan 69^o tan 79^o tan 89^o = cot 89^o cot 79^o cot 69^o tan 69^o tan 79^o tan 89^o = cot 89^o cot 79^o cot 69^o _{= 1}

Q11.

Solution

Q12. In figure, AC = 13 cm, BC = 12 cm, then sec equals :

Solution

Q13. If 6 cot + 2 cosec = cot + 5 cosec, then cos is:

Solution

6 cot + 2 cosec = cot + 5 cosec 5 cot = 3 cosec

Q14. In figure, if D is mid-point of BC, the value of _{$fraction numerator cot space straight y degree over denominator cot space straight x degree end fraction$}is:

1)
2)
3) 2
4)

Solution

Q15. If tan 2A = cot (A - 18^o), then the value of A is

1) 27^o
2) 24^o
3) 36^o
4) 18^o

Solution

Q16. Find the length of the diagonals of the rhombus shown in the figure given below.

Solution

We know that the diagonals of a rhombus bisect each other at right angles. In AOD,

Q17. In the given figure find tan A - cot C.

Solution

Q18. The value of _{$fraction numerator cot space 45 degree over denominator sin space 30 degree plus space cos space 60 degree end fraction$}is equal to:

1)
2) 1
3)
4)

Solution

Q19. Prove that:.

Solution

= 1 + tan² = sec²

Q20. [cos⁴ A - sin⁴A] is equal to:

1) 2 cos² A + 1
2) 2 sin² A - 1
3) 2 sin² A + 1
4) 2 cos² A - 1

Solution

cos⁴ A - sin⁴A = cos⁴ A - (1 - cos²A)² = cos⁴ A - (1 + cos⁴A - 2 cos²A) = 2 cos² A - 1

Q21. Prove that

Solution

Hence, R.H.S. = L.H.S.

Q22. If sin (A - B) = and cos (A + B) = , then the value of B is:

1) 15^o
2) 60^o
3) 45^o
4) 0^o

Solution

Q23.

Solution

Q24. If tan = 3 sin, prove that sin² - cos² =.

Solution

LHS = sin² - cos² = 1 - 2 cos²_{[since
sin2 = 1 - cos2]} =1 - 2 _{= RHS}

Q25. If x = 2 , y = 2

cos squared straight theta

+ 1, then the value of x + y is:

1) 1
2) 2
3) 3
4)

Solution

straight x plus straight y equals 2 sin squared straight theta space plus 2 cos squared straight theta plus 1 straight x plus straight y equals 2 plus 1 equals 3

Q26. If cos x = cos 60^o cos 30^o + sin 60^o sin 30^o, then the value of x is

1) 60^o
2) 15^o
3) 30^o
4) 45^o

Solution

cos x = cos 60^o cos 30^o + sin 60^o sin 30^o We know cos 30^o = Hence, x = 30^o

Q27.

open parentheses fraction numerator cos space straight A over denominator cot space straight A end fraction plus sin space straight A close parentheses space is

1) sec A
2) 2 cos A
3) 2 sin A
4) cot A

Solution

fraction numerator cos space straight A over denominator cot space straight A end fraction plus sin space straight A equals space fraction numerator cos begin display style space end style begin display style straight A end style over denominator begin display style fraction numerator cos begin display style space end style begin display style straight A end style over denominator sin begin display style space end style begin display style straight A end style end fraction end style end fraction plus sin space straight A equals sin space straight A space plus space sin space straight A equals 2 sin space straight A

Q28. Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A.

Solution

LHS = (sin A + cosec A)² + (cos A + sec A)² = sin²A + cosec²A + 2 sin A cosec A + cos²A + sec²A + 2cos A sec A = 1 + 2 + 2 + cosec² A + sec²A = 1 + 2 + 2 + cot² A + 1 + 1 + tan²A = 7 + cot²A + tan²A = RHS

Q29. sin (60^o + ) - cos (30^o - ) is equal to (where (60^o + ) and (30^o - ) are both acute angles):

1) 1
2) 2 sin
3) 2 cos
4) 0

Solution

sin (60^o + ) - cos (30^o - ) = sin (60^o + ) - cos (90^o - (60^o + )) = sin (60^o + ) - sin (60^o + ) = 0

Q30. Prove that 1 +

Solution

LHS = = 1 + = 1 + = 1 + cosec - 1 = cosec = = RHS

Q31. If sin (60^o - ) = 1, then is:

1) 45^o
2) 30^o
3) 60^o
4) 15^o

Solution

sin (60^o - ) = 1 sin (60^o - ) = _{$fraction numerator 1 over denominator square root of 2 end fraction$}= sin 45^o 60^o - = 45^o = 15^o

Q32. Prove that

Solution

Q33.

Solution

Q34. If tan + sin = m and tan - sin = n, show that m² - n² = 4

Solution

m² - n² = (m - n) (m + n) = (2 sin ) (2 tan ) = 4 sin tan ... (i) mn = (tan + sin ) (tan - sin ) = tan² - sin² ... (ii) (i) and (ii) m² - n² = 4 .

Q35. Prove that cos⁸ - sin⁸ = (cos² - sin²) (1 - 2sin² cos²)

Solution

Q36. The value of is

1) sin A.cos A
2) 1-sin A.cos A
3) 1+sin A.cos A
4) tan A

Solution

=(1+ sin A.cos A)[]

Q37. If A, B, C are interior angles of ABC, show that:

Solution

cosec² - tan² = cosec² = sec² = 1

Q38. Prove: = cosec

+ cot

Solution

L.H.S. =

Q39. If x = r sin A cos C, y = r sin A sin C, z = r cos A, prove that r² = x² + y² + z².

Solution

x = r sinA cosC, y = r sinA sinC, z = r cosA x² + y² + z² = r²sin²A cos²C + r²sin²A sin²C + r²cos²A = r²sin²A (cos²C + sin²C) + r²cos²A = r²sin²A + r²cos²A = r²(sin²A + cos²A) = r² (since, sin²+ cos² = 1)

Q40. If sin = , find the value of

Solution

sin ₌ tan =

Q41. In figure below, ABC is right-angled at B and tan A =

4 over 3

. If AC = 15 cm, then the length of AB is :

1) 9 cm
2) 4 cm
3) 12 cm
4) 3 cm

Solution

_{$tan space straight A space equals BC over AB BC over AB equals 4 over 3 Let space BC equals space 4 space straight k space and space AB equals 3 space straight k AC squared equals AB squared plus BC squared 225 equals 9 space straight k squared plus 16 space straight k squared straight k squared equals 9 straight k equals 3 rightwards double arrow AB equals 3 cross times 3 equals 9$}

Q42. In the figure given below, ACB = 90^o, BDC = 90^o, CD = 4 cm, BD = 3 cm, AC = 12 cm. cos A - sin A is equal to:

Solution

Applying Pythagoras theorem, BC² = CD² + BD² = (4 cm)² + (3 cm)² BC = 5 cm AB² = AC² + BC² = (12 cm)² + (5 cm)² AB = 13 cm cos A - sin A

Q43. In the figure, if PS = 14 cm, the value of tan a is equal to:

Solution

tan a = … (1) ST = PS - TP = 14 cm - 5 cm = 9 cm Using Pythagoras theorem in triangle PQR, we have: (13)² = PQ² + (5)² PQ² = 169 - 25 = 144 PQ = 12 cm TR = PQ = 12 cm From (1), tan a =

12 over 9

Q44. If x = a cos + b sin and y = b cos - a sin , then prove that x² + y² = a² + b².

Solution

x = a cos + b sin ...(1) y = b cos - a sin ...(2) Squaring and adding, we get, x² + y² = a² cos² + b² sin² + 2ab sin cos + b² cos² + a² sin² - 2ab sin cos = a² (sin² + cos²) + b²(sin² + cos²) = a² + b² Hence, proved

Q45. The value of cos θ cos(90° - θ) - sin θ sin (90° - θ) is:

1) 1
2) 2
3) -1
4) 0

Solution

cos θ cos(90° - θ) - sin θ sin (90° - θ) = cos θ sin θ - sin θ cos θ = 0

Q46. If sin = cos, then value of is :

1) 0^o
2) 90^O
3) 30^o
4) 45^o

Solution

Given,

_{We know $sin 45 degree space equals fraction numerator 1 over denominator square root of 2 end fraction space equals space cos 45 degree space$} Thus,

Q47. The value of 5 tan²

straight theta

- 5 sec²

straight theta

is:

1) 5
2) -5
3) 0
4) 1

Solution

5 tan² - 5 sec²=5 tan² - 5 (1 + tan²) = 5 tan² - 5 - 5 tan² = -5

Q48. If sin θ + cos θ = sin (90° - θ), then find the value of tan θ.

Solution

begin mathsize 12px style sinθ space plus space cosθ space equals space square root of 2 space sin open parentheses 90 degree minus straight theta close parentheses sinθ space plus space cosθ space equals space space square root of 2 cosθ space sinθ space equals space space square root of 2 cosθ space minus space space cosθ sinθ space equals space cosθ space open parentheses square root of 2 minus 1 close parentheses fraction numerator sinθ over denominator cosθ space end fraction equals square root of 2 minus 1 tanθ space equals space square root of 2 minus 1 end style

Q49. [(sec A + tan A) (1 - sin A)] on simplification gives:

1) sec² A
2) sin A
3) tan²A
4) cos A

Solution

(sec A + tan A) (1 - sin A)

Q50. Without using trigonometric tables, prove that

fraction numerator size 12px cos to the power of size 12px 2 size 12px 49 size 12px degree size 12px space size 12px plus size 12px space size 12px cos to the power of size 12px 2 size 12px 41 size 12px degree over denominator size 12px sin to the power of size 12px 2 size 12px 31 size 12px degree size 12px space size 12px plus size 12px space size 12px sin to the power of size 12px 2 size 12px 59 size 12px degree end fraction size 12px space size 12px plus size 12px space size 12px 2 size 12px tan size 12px 35 size 12px degree size 12px space size 12px tan size 12px 55 size 12px degree size 12px equals size 12px 3

Solution

fraction numerator size 12px cos to the power of size 12px 2 size 12px 49 size 12px degree size 12px space size 12px plus size 12px space size 12px cos to the power of size 12px 2 size 12px 41 size 12px degree over denominator size 12px sin to the power of size 12px 2 size 12px 31 size 12px degree size 12px space size 12px plus size 12px space size 12px sin to the power of size 12px 2 size 12px 59 size 12px degree end fraction size 12px space size 12px plus size 12px space size 12px 2 size 12px tan size 12px 35 size 12px degree size 12px space size 12px tan size 12px 55 size 12px degree size 12px equals size 12px 3

= 1 + 2 1 = 1 + 2 = 3

Q51. From the figure, the value of cosec A + cot A is :

Solution

Q52. If cot A = _{$12 over 5$}, then the value of (sin A + cos A) cosec A is :

1) 1
2)
3)
4)

Solution

open parentheses sin space straight A space plus space cos space straight A close parentheses cross times cosec space straight A sin space straight A space cross times cosec space straight A space plus space cos space straight A space cross times cosec space straight A equals 1 plus fraction numerator cos space straight A over denominator sin space straight A end fraction equals 1 plus cot space straight A equals 1 plus 12 over 5 equals 17 over 5

Q53. If cosec²(1 + cos) (1 - cos) = , then the value of is:

1) cos²
2) -1
3) 0
4) 1

Solution

cosec squared straight theta space open parentheses 1 plus cos space straight theta close parentheses open parentheses 1 minus cos space straight theta close parentheses cosec squared straight theta space open parentheses 1 minus cos squared space straight theta close parentheses cosec squared straight theta space sin squared space straight theta

Q54. The maximum value of is:

1) 0
2)
3)
4) 1

Solution

We know and the maximum value of sin is 1. Hence, the maximum value of is 1.

Q55. If A, B, C are interior angles of ABC, show that:

Solution

= = 1

Q56. If cos (40^o + A) = sin 30^o, the value of A is:

1) 40^o
2) 20^o
3) 30^o
4) 60^o

Solution

cos (40^o + A) = sin 30^o cos (40^o + A) = sin (90 - 60^o) cos (40^o + A) = cos 60^o 40^o + A = 60^o A = 20^o

Q57.

Solution

Q58. Without using trigonometric tables, prove that:

Solution

= 1 + 1 = 2

Q59. Given that sin A = and cos B = then the value of A + B is:

1) 15^o
2) 30^o
3) 75^o
4) 45^o

Solution

Q60. ABC is right angled at A, the value of tan B tan C is:

1) 0
2) -1
3) None of the above
4) 1

Solution

tan space straight B space cross times space tan space straight C equals AC over AB cross times AB over AC equals 1

Q61. If cosec = x + , prove that: cosec + cot = 2x or

Solution

We know: 1 + cot² = cosec² cot² = cosec² - 1 cot² = = x² + = x² + = cot = cot = x - or cot = -x + cosec + cot = 2x or cosec + cot x =

Q62.

begin mathsize 12px style If space cos open parentheses straight alpha space plus space straight beta close parentheses space equals space 0 comma space then space sin open parentheses straight alpha space minus space straight beta close parentheses space can space be space reduced space to space colon end style

1) $sin space 2 straight alpha$
2) $cos space straight beta$
3) $sin space straight alpha$
4) $cos space 2 straight beta$

Solution

begin mathsize 12px style cos open parentheses straight alpha space plus space straight beta close parentheses space equals space 0 straight alpha space plus space straight beta space equals space 90 degree straight alpha space equals space 90 degree space minus space straight beta space sin open parentheses straight alpha space minus space straight beta close parentheses space equals space sin open parentheses 90 degree space minus space straight beta space minus space straight beta close parentheses equals space sin open parentheses 90 degree space minus space 2 straight beta close parentheses equals space cos 2 straight beta end style

Q63. Prove the following trigonometric identities.

square root of fraction numerator 1 minus cosA over denominator 1 plus cosA end fraction end root plus square root of fraction numerator 1 plus cosA over denominator 1 minus cosA end fraction end root equals 2 cosecA

Solution

table attributes columnalign left end attributes row cell Consider text   the   LHS ,  end text square root of fraction numerator 1 minus cosA over denominator 1 plus cosA end fraction end root plus square root of fraction numerator 1 plus cosA over denominator 1 minus cosA end fraction end root colon end cell row cell square root of fraction numerator 1 minus cosA over denominator 1 plus cosA end fraction end root plus square root of fraction numerator 1 plus cosA over denominator 1 minus cosA end fraction end root equals fraction numerator open parentheses square root of 1 minus cosA end root close parentheses squared plus open parentheses square root of 1 plus cosA end root close parentheses squared over denominator square root of 1 plus cosA end root square root of 1 minus cosA end root end fraction end cell row cell text                                                end text equals fraction numerator 1 minus cosA plus 1 plus cosA over denominator square root of 1 minus cos squared straight A end root end fraction end cell row cell text                                                end text equals fraction numerator 2 over denominator square root of sin squared straight A end root end fraction end cell row cell text                                                end text equals 2 over sinA end cell row cell text                                                end text equals 2 cosecA end cell row cell text                                                end text equals RHS end cell row cell Hence text   proved. end text end cell end table

Q64. Prove that sin A(1 + tan A) + cos A (1 + cot A) = sec A + cosec A.

Solution

LHS = sin A(1 + tan A) + cos A (1 + cot A) = sin A

Q65. Prove that

Solution

Q66. If sin (20^o + ) = cos 30^o, then the value of is:

1) 30^o
2) 50^o
3) 40^o
4) 20^o

Solution

sin (20^o + ) = cos 30^o We know cos 30^o = sin (20^o + ) = = sin 60^o 20^o + = 60^o = 40^o

Q67. Find the value of tan 60^o geometrically.

Solution

Consider an equilateral ABC and let a be the length of each side of ABC. AB = BC = CA = a

Draw AD BC ABD ACD BD = DC BD = BC = a _andBAD = CAD BAD = CAD = 30^o _{Using Pythagoras theorem in ABD,} _{AD² = AB² - BD²}

Q68. Given that cos =

straight m over straight n

then tan is equal to

Solution

Q69. In figure below sec

alpha

is:

Solution

_{Using Pythagoras theorem in ABC,} _{Using Pythagoras theorem in ACD,}

Q70. Prove that : (sin + cosec)² + (cos + sec)²= 7 + tan² + cot²

Solution

LHS = (sin + cosec)² + (cos + sec)² = sin² + cosec² + 2sin cosec + cos² + sec² + 2cos sec = sin² + cos² + 2 + 2 + cosec² + sec² = 1 + 4 + (1 + cot²) + (1 + tan²) = 7 + tan² + cot² = RHS

Q71. In ABC, right angled at B, if cot A= then the value of cos A sin C + sin A cos C

1)
2) 1
3)
4) 3

Solution

Given that, cot A = =

fraction numerator square root of 3 over denominator 1 end fraction

If BC = k then AB = k By Pythagoras theorem, we have: AC² = AB² + BC² = (k)² + (k)² AC² = 3k² + k² = 4k² Thus, AC = 2k Hence, sin A =

1 half

and cos A =

fraction numerator square root of 3 over denominator 2 end fraction

Also, sin C =

fraction numerator square root of 3 over denominator 2 end fraction

and cos C =

1 half

Therefore, cos A sin C + sin A cos C =

Q72. If cos A + cos² A = 1, then sin² A + sin⁴ A is

1) -1
2) 1
3) 2
4) 0

Solution

Q73. If and f are the acute angles of a right triangle, and

Solution

The two angles and f being the acute angles of a right triangle must be complementary angles. So, Substituting, in above equation,

Q74.

Solution

Q75. Given 15cot A = 8, find sin A and sec A.

Solution

Q76. Prove that

Solution

Q77. If sin + sin² = 1, then find the value of cos² + cos⁴.

Solution

sin + sin² = 1 sin = 1 - sin² = cos² sin² = cos⁴ 1 - cos² = cos⁴ cos² + cos⁴ = 1

Q78. Prove that: cos sin -

Solution

LHS = cos sin - = cos sin - = cos sin - sin³ cos - cos³ sin = cos sin - sin cos (sin² + cos²) = cos sin - sin cos1 [Since, sin² + cos² = 1] = 0 = RHS

Q79. If x = 3 sec² - 1, y = tan² - 2 then x - 3y is equal to

1) 4
2) 3
3) 8
4) 5

Solution

x = 3 sec² - 1, y = tan²- 2 x - 3y = 3 sec² - 1 - 3tan²+ 6 ₌3(1) + 5 = 8 (Since, sec² - tan² = 1)

Q80. Prove that:

Solution

= 1 = RHS

Q81. If x. tan 45^o. cot60^o = sin 30^o. cosec60^o, then the value of x is:

1) 1
2)
3)
4)

Solution

x. tan 45^o. cot60^o = sin 30^o. cosec60^o

Q82.

Solution

Q83. ABC is a right triangle right angled at C, then the value cosec²A - tan²B is :

1) -1
2) -2
3) 2
4) 1

Solution

cosec²A - tan²B

Q84. If 7 sin² q + 3 cos² q = 4, then prove that sec q + cosec q = 2 +.

Solution

7 sin² + 3 cos² = 4 7 sin² + 3 (1 - sin² ) = 4 sin = sec 30^o + cosec30^o =

Q85. If 5 tan = 4, find the value of

Solution

5 tan = 4 5 sin = 4 cos

Q86. Prove that:

Solution

Q87. Prove that (cosec A - sin A) (sec A - cos A) =

Solution

LHS = (cosec A - sin A) (sec A - cos A) RHS = Hence, LHS = RHS

Q88. If sin (A + B) = cos (A - B) = and A,B (A >B) are acute angles, find the value of A and B.

Solution

Given, sin (A + B) = cos (A - B) = Therefore, A + B = 60^o ; A - B = 30^o Solving for A and B, we get, A = 45^o and B = 15^o

Q89. If tan + sin = m and tan - sin = n Show that (m² - n²)² = 16 mn

Solution

tan + sin = m tan - sin = n (m + n) (m - n) = 2 tan 2 sin m² - n² = 4 tan sin L.H.S. = (m² - n²)² = 16 tan² sin² R.H.S. = 16 mn = 16 (tan + sin ) (tan - sin ) = 16 (tan² - sin² ) = 16 = 16 = 16 = 16 tan² sin² L.H.S. = R.H.S. (m² - n²)² = 16mn

Q90. If tan = 3 sin , find the value of sin² - cos²

Solution

tan = 3 sin

Q91. If A, B, C are the interior angles of ΔABC, then prove that _{$Error converting from MathML to accessible text.$}.

Solution

begin mathsize 12px style Since space straight A space plus space straight B space plus space straight C space equals space 180 degree straight A space plus space straight B space equals space 180 degree space minus space straight C comma Now comma space straight L. straight H. straight S equals cos open parentheses fraction numerator straight A plus straight B over denominator 2 end fraction close parentheses equals cos open parentheses fraction numerator 180 degree minus straight C over denominator 2 end fraction close parentheses equals cos open parentheses 90 minus straight C over 2 close parentheses equals sin straight C over 2 equals straight R. straight H. straight S. end style

Q92. If = n then Show that (m² + n²) cos² = n².

Solution

Q93. If 3 cot A = 4, find the value of

Solution

Q94. If cos =

1 half

, then the value of cos [cos - sec] is:

Solution

cos = _{$1 half$}

Q95. Evaluate:

Solution

Q96. Prove that

Solution

LHS =

= 1 + cos A

= RHS

Q97. In fig., PQR is right angled at Q, QR = 6 cm QPR = 60^o. Find the length of PQ and PR.

Solution

Q98.

Solution

Q99. Prove that

Solution

Q100. In triangle ABC, right angled at A, if AB = 5, AC = 12 and BC = 13, find sin B, cos C and tan B.

Solution

Q101. Evaluate:

Solution

Q102. If sin =

1 half

, then the value of sin (sin - cosec) is

Solution

sin space straight theta space open parentheses sin space straight theta space minus space cosec space straight theta close parentheses equals 1 half open parentheses 1 half minus 2 close parentheses equals fraction numerator negative 3 over denominator 4 end fraction

Q103. Prove that

Solution

L.H.S =

Q104. Prove:

Solution

Q105. If a cot θ+ b cosec θ = p and b cot θ + a cosec θ= q then p² - q² is equal to

1) a²- b²
2) a² + b²
3) b² + a²
4) b² - a²

Solution

p = a cot θ + b cosec θ q = b cot θ + a cosec θ p² - q² = (a cot θ + b cosec θ)² - (b cot θ + a cosec θ)² = a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ - b² cot² θ - a² cosec² θ - 2ab cot θ cosec θ = a² (cot² θ - cosec²θ) + b² (cosec² θ - cot² θ) = b² - a² (Since, cosec² A - cot² A = 1)

Q106.

Solution

Q107. If tan (A + B) = and tan (A - B) = , 0^o < A + B 90^o; A > B find A and B.

Solution

Given, tan (A + B) = and tan (A - B) = Therefore, A + B = 60^o; A - B = 30^o Solving the two equations, we get, A = 45^o ; B = 15^o

Q108. Prove that tan² + cot² + 2 = sec² cosec²

Solution

= cosec² sec² = RHS

Q109. If cos - sin = sin, prove that cos + sin = cos .

Solution

Given cos - sin = sin

Q110. If sec 4A = cosec (A - 20^o) where 4 A is an acute angle, fine the value of A.

Solution

Given, sec 4A = cosec (A - 20^o) Since, sec 4 A = cosec (90^o - 4 A) (90^o - 4 A) and (A - 20^o) are acute angles 90^o - 4 A = A - 20^o A = 22^o

Q111. Prove that:

Solution

Q112.

Solution

Q113. Prove that:

Solution

Q114. If a cos - b sin = c, prove that (a sin + b cos ) = .

Solution

(a cos - b sin )² = c² a² cos² + b² sin² - 2 ab sin . cos = c² a² (1 - sin²) + b²(1 - cos²) - 2 ab sin . cos = c² a² + b² - c² = (a sin + b cos )² (a sin + b cos ) =

Q115.

Solution

Q116. Prove that .

Solution

Q117. _Evaluate:

Solution

begin mathsize 12px style fraction numerator 2 sin 68 degree over denominator cos 22 degree end fraction minus 2 fraction numerator begin display style tan end style open parentheses 90 degree minus 15 degree close parentheses over denominator 5 cot 15 degree end fraction minus fraction numerator 3 tan 45 degree space tan 20 degree space tan 40 degree space tan 50 degree space tan 70 degree over denominator 5 open parentheses sin squared 70 degree space plus space sin squared 70 degree close parentheses end fraction equals fraction numerator 2 cos open parentheses 90 degree minus 68 degree close parentheses over denominator cos begin display style 22 end style begin display style degree end style end fraction minus fraction numerator 2 cot 15 degree over denominator 5 cot 15 degree end fraction minus fraction numerator 3 cross times 1 cross times cot open parentheses 90 degree minus 20 degree close parentheses cot open parentheses 90 degree minus 40 degree close parentheses tan 50 degree tan 70 degree over denominator 5 open curly brackets cos squared open parentheses 90 degree minus 20 degree close parentheses space plus space space sin squared 20 degree close curly brackets end fraction equals fraction numerator 2 cos 22 degree over denominator cos 22 degree end fraction minus 2 over 5 minus fraction numerator 3 cot 70 degree space cot 50 degree space tan 50 degree tan 70 degree over denominator 5 open parentheses space cos squared 20 degree plus sin squared 20 degree close parentheses end fraction equals 2 minus 2 over 5 minus 3 over 5 equals fraction numerator 10 minus 2 minus 3 over denominator 5 end fraction equals 5 over 5 equals 1 end style

Q118. If sin + cos = m and sec+cosec = n, prove that n ( = 2m.

Solution

Given: sin + cos = m and sec + cosec = n Consider L.H.S. = n = = =

Q119. Prove that

Solution

L.H.S.

Q120. Evaluate cos 30^o cos 45^o - sin 30^o sin 45^o.

Solution

Q121. In , right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find the value of sin P.

Solution

Given, PR + QR = 25 cm and PQ = 5 cm In right PQR, PR² = PQ² + QR² (by Pythagoras theorem) (25 − QR)² = 5² + QR² QR = 12 PR + QR = 25 PR = 13 sin P =

Q122. Prove that

= 2 sec

Solution

LHS =

Q123. If A + B = 90^o, then prove that

Solution

Q124. Prove that:

Solution

(since, sin²A + cos²A = 1)

Q125. Evaluate:

begin mathsize 12px style fraction numerator sec squared space left parenthesis 90 degree minus straight theta right parenthesis minus cot squared space straight theta over denominator 2 left parenthesis sin squared space 25 degree plus sin squared space 65 degree right parenthesis end fraction plus fraction numerator 2 cos squared space 60 degree space tan squared space 28 degree space tan squared space 62 degree over denominator 3 left parenthesis sec squared space 43 degree minus cot squared space 47 degree right parenthesis end fraction plus fraction numerator cot space 40 degree over denominator tan space 50 degree end fraction end style

Solution

begin mathsize 12px style fraction numerator sec squared space left parenthesis 90 degree minus straight theta right parenthesis minus cot squared space straight theta over denominator 2 left parenthesis sin squared space 25 degree plus sin squared space 65 degree right parenthesis end fraction plus fraction numerator 2 cos squared space 60 degree space tan squared space 28 degree space tan squared space 62 degree over denominator 3 left parenthesis sec squared space 43 degree minus cot squared space 47 degree right parenthesis end fraction plus fraction numerator cot space 40 degree over denominator tan space 50 degree end fraction equals fraction numerator cosec squared space straight theta minus cot squared space straight theta over denominator 2 left square bracket sin squared space 25 degree plus sin squared space left parenthesis 90 degree minus 25 degree right parenthesis right square bracket end fraction plus fraction numerator 2 cos squared space 60 degree space tan squared space 28 degree space tan squared space left parenthesis 90 degree minus 28 degree right parenthesis over denominator 3 left square bracket sec squared space 43 degree minus cot squared space left parenthesis 90 degree minus 43 degree right parenthesis right square bracket end fraction plus fraction numerator cot space 40 degree over denominator tan space left parenthesis 90 degree minus 40 degree right parenthesis end fraction equals fraction numerator 1 over denominator 2 left parenthesis 1 right parenthesis end fraction plus fraction numerator 2 cross times open parentheses begin display style 1 half end style close parentheses squared space tan squared space 28 degree space cot squared space 28 degree over denominator 3 left square bracket sec squared space 43 degree minus tan squared space 43 degree right square bracket end fraction plus fraction numerator cot space 40 degree over denominator cot space 40 degree end fraction equals 1 half plus fraction numerator 2 cross times begin display style 1 fourth end style over denominator 3 left parenthesis 1 right parenthesis end fraction plus 1 equals 1 half plus 1 over 6 plus 1 equals fraction numerator 3 plus 1 plus 6 over denominator 6 end fraction equals 10 over 6 equals 5 over 3 end style

Q126. Prove the following identity: (sin + sec )² + (cos + cosec )² = (1 + sec cosec )².

Solution

Consider L.H.S. = (sin + sec )² + (cos + cosec )² = sin² + sec² + 2 tan + cos² + cosec² + 2 cot = 1 + sec² cosec² + 2 sec cosec = (1 + sec cosec )² = R.H.S

Q127.