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Trigonometric Ratios Of Complementary Angle

Class 10th Mathematics West Bengal Board Solution
Let Us Work Out 24
  1. sin38^circle /cos52^circle Let us evaluate :
  2. cosec79^circle /sec11^circle Let us evaluate :
  3. tan27^circle /cot63^circle Let us evaluate :
  4. sin66° - cos24° = 0 Let us show that:
  5. cos^2 57° + cos^2 33° = 1 Let us show that:
  6. cos^2 75° - sin^2 15° = 0 Let us show that:
  7. cosec^2 48° - tan^2 42° = 1 Let us show that:
  8. sec70°sin20° + cos20°cosec70° = 2 Let us show that:
  9. sin^2 α + sin^2 β = 1 If two angles and are complementary angle, let us show that…
  10. cotβ + cosβ = If two angles and are complementary angle, let us show that…
  11. secalpha /cosalpha -cot^2beta = 1 If two angles and are complementary angle, let us…
  12. If sin 17° = x/y , let us show that sec 10° - sin73° = x^2/y root y^2 - x^2…
  13. Let us show that sec^2 12° - 1/tan^278^circle = 1
  14. ∠A + ∠B = 90°, let us show that 1 + tana/tanb = sec^2a
  15. Let us show that cosec^2 22°cot^2 68° = sin^2 22° + sin^2 68° + cot^2 68°…
  16. If ∠A + ∠B = 90°, let us show that, root sinp/cosq-sinpcosq - cosp…
  17. Let us prove that cot 12°cot38° cot ° cot52° cot 78° cot 60° = 1/root 3…
  18. AOB is a diameter of a circle with centre O and C is any point on the circle, joining…
  19. ABCD is a rectangular figure, joining A,C let us prove that (i) tan∠ACD = cot∠ACB (ii)…
  20. Q12A1 The value of (sin 43° cos47° + cos43° sin47°)A. 0 B. 1 C. sin4° D. cos4°…
  21. Q12A2 The value of (tan35^circle /cot55^circle + cot78^circle /tan12^circle) isA. 0 B. 1…
  22. Q12A3 The value of {cos(40° + θ) - sin(50° - θ)} isA. 2cosθ B. 7sinθ C. 0 D. 1…
  23. Q12A4 ABC is triangle. Sin (b+c/2) = A. sin a/2 B. cos a/2 C. sinA D. cosA…
  24. Q12A5 If A + B = 90° and tan A = 3/4 , value of cot B isA. 3/4 B. 4/3 C. 3/5 D. 4/5…
  25. Let us write whether the following statements are true of false: (i) The value of…
  26. Let us fill up the blanks: (i) The value of (tan 15° x tan 45° x tan 60° tan 75°)…
  27. If sin 10θ = cos8θ and 10θ is a positive acute angle, let us find the value of tan…
  28. If tan 4θ × tan 6θ = 1 and 6θ is a positive acute angle, let us find the value of θ.…
  29. let us find the value of 2sin^263^circle + 1+2sin^227^circle /3cos^217^circle -…
  30. let us find the value of (tan 1° × tan2° × tan 3°………..tan89°)
  31. If sec 5A = cosec (A + 36°) and 5A is a positive acute angle, let us find the value…
Let Us Work Out 24
Question 1.

Let us evaluate :
Answer:

Given,


Need to evaluate the given equation


⇒ we know that cos(90 - θ) = sinθ


∴ cos 52° = cos(90 - 38)°


= sin 38°


⇒


= 1
Question 2.

Let us evaluate :
Answer:

Given,


Need to evaluate the given equation


⇒ we know that sec(90 - θ) = cosecθ


∴ sec 11° = sec(90 - 79)°


= cosec 79°


⇒


= 1
Question 3.

Let us evaluate :
Answer:

Given,

Need to evaluate the given equation


⇒ we know that tan(90 - θ) = cotθ


∴ tan 27° = tan(90 - 63)°


= cot 63°


⇒


= 1
Question 4.

Let us show that:

sin66° - cos24° = 0
Answer:

Given, sin66° - cos24° = 0


Need to prove the given equation as zero


⇒ we know that cos(90 - θ) = sinθ


∴ cos24° = cos(90 - 66)°


= sin66° - - - eq (1)


⇒ sin66° - cos24° = 0


[substitute eq(1)]


⇒ sin66° - sin66° = 0


Hence, LHS = RHS
Question 5.

Let us show that:

cos257° + cos233° = 1
Answer:

Given, cos257° + cos2 33°


Need to prove the given equation as one


⇒ we know that cos(90 - θ) = sinθ


∴ cos33° = cos(90 - 57)°


= sin57° - - - eq (1)


⇒ Substitute eq(1) in the given equation


∴ cos257° + sin257°


⇒ we know that, sin2θ + cos2θ = 1


∴ cos257° + sin257° = 1


Hence, proved
Question 6.

Let us show that:

cos2 75° - sin215° = 0
Answer:

Given, cos275° - sin215°


Need to prove the given equation as zero


⇒ we know that cos(90 - θ) = sinθ


∴ cos75° = cos(90 - 15)°


= sin15° - - - eq (1)


⇒ Substitute eq(1) in the given equation


∴ cos275° - sin215° =


sin215° - sin215° = 0


Hence, proved
Question 7.

Let us show that:

cosec248° - tan242° = 1
Answer:

Given, cosec2 48° - tan2 42°


Need to prove the given equation as one


⇒ we know that tan(90 - θ) = cotθ


∴ tan42° = tan(90 - 48)°


= cot48° - - - eq (1)


⇒ Substitute eq(1) in the given equation


∴ cosec2 48° - cot248°


⇒ we know that


And


∴


=


[ sin2θ + cos2θ = 1


∴ 1 - cos2θ = sin2θ]


⇒


= 1


∴ cosec2 48° - tan2 42° = 1


Hence, proved
Question 8.

Let us show that:

sec70°sin20° + cos20°cosec70° = 2
Answer:

Given, sec70°sin20° + cos20°cosec70° = 2


Need to prove the given equation as two


⇒ we know that sec(90 - θ) = cosecθ and cosec(90 - θ) = secθ


∴ sec70° = sec(90 - 20)°


= cosec20° - - - eq (1)


And cosec70° = cosec(90 - 20)


= sec20° - - - - - eq(2)


⇒ Substitute eq(1) and eq(2) in the given equation


∴ cosec20°sin20° + cos20°sec20° = 2


⇒


[ and ]


⇒ = 2


⇒ 2 = 2


Hence, proved
Question 9.

If two angles and are complementary angle, let us show that

sin2α + sin2β = 1
Answer:

Given, two angles are complementary = 90°


⇒ sin2α + sin2β


⇒ sin2α + sin2(90 - α)


⇒ sin2α + cos2α


= 1


[sin2θ + cos2θ = 1]
Question 10.

If two angles and are complementary angle, let us show that

cotβ + cosβ =


Answer:

Given, cotβ + cosβ


[


⇒


= cosβ()


= cosβ( )


=


=


=



Question 11.

If two angles and are complementary angle, let us show that
Answer:

Given,


⇒


⇒ [ and ]


⇒


⇒


[cos α = cos(90 - β)]


⇒


⇒


⇒


[sin2θ + cos2θ = 1]


⇒


= 1
Question 12.

If sin 17° = , let us show that sec 10° - sin73° =


Answer:

Given,


To show,


[sec θ =


⇒


⇒ sin73° = sin(90 - 17) = cos17


⇒


=


[sin2θ + cos2θ = 1]


=


=


=


=


=



Question 13.

Let us show that sec212° -


Answer:

Given, = 1


[sec =


⇒ tan78 = tan(90 - 12) = cot12


∴ = 1


⇒


=


=


=


= 1


Hence, proved
Question 14.

∠A + ∠B = 90°, let us show that


Answer:

Given, ∠A + ∠B = 90°


To show that


⇒


⇒ tanB = tan(90 - A) = cotA


=


[cotA = ]


=


= 1 + tan2A


= sec2A
Question 15.

Let us show that cosec222°cot268° = sin222° + sin268° + cot268°
Answer:

Given, cosec222°cot268° = sin222° + sin268° + cot268°


⇒ sin222° + sin268° + cot268°


= sin222° + sin2(90 - 22)° + cot268°


= sin222° + cos222° + cot268°


= 1 + cot268°


= cosec268°
Question 16.

If ∠A + ∠B = 90°, let us show that,


Answer:

Given, ∠ A + ∠ B = 90°


To show = cosP


⇒


=


=


=


=


= sinQ


= sin(90 - p)


= cosP


Hence proved
Question 17.

Let us prove that cot 12°cot38° cot ° cot52° cot 78° cot 60° =


Answer:

Given, cot12° cot38° cot52° cot78° cot60°


⇒ we know that cot60 =


⇒ cot12° cot38° cot52° cot78° ()


= cot(90 - 78)cot(90 - 52)cot52° cot78° ()


= tan78° tan52° cot52° cot78° ()


= tan78° tan52° ()


= ()
Question 18.

AOB is a diameter of a circle with centre O and C is any point on the circle, joining A.C; B,C; and O, C let us show that

(i) tan∠ABC = cot ∠ACO

(ii) sin2∠BCO + sin2∠ACO = 1

(iii) cosec2∠CAB - 1 = tan2∠ABC
Answer:


(i) tan∠ABC = cot ∠ACO


⇒ ∠ AOC = 90° = ∠ BOC


⇒ ∠ CAO = 60° and


∠ ACO = 30°


⇒ tan∠ ABC = tan(90 - ∠ ACO)


[tan(90 - θ) = cotθ ]


= cot∠ ACO


(ii) sin2∠BCO + sin2∠ACO = 1


⇒ sin2∠BCO + sin2∠ACO = 1


⇒ sin2∠BCO + sin2∠ACO


⇒ sin2∠BCO + sin2(90 - ∠ BCO)


⇒ sin2∠BCO + cos2∠BCO


= 1


[since, sin2θ + cos2θ = 1]


(iii) cosec2∠CAB - 1 = tan2∠ABC


⇒ cosec2∠CAB - 1


⇒ cosec(90 - ∠CAB) = sec∠CAB


⇒ ∠ CAB = ∠ ABC


⇒ sec2∠CAB - 1 = tan2∠ ABC


⇒ sec2∠ ABC – 1 = tan2∠ ABC


⇒ tan2∠ ABC = tan2∠ ABC
Question 19.

ABCD is a rectangular figure, joining A,C let us prove that

(i) tan∠ACD = cot∠ACB

(ii) tan2∠CAD + 1 =


Answer:


AC is the diagonal line joining ABCD


(i) tan∠ACD = cot∠ACB


⇒ ∠ACD + ∠ACB = 90


⇒ tan∠ACD


= tan(90 - ∠ACB)


= cot∠ACB


(ii) tan2∠CAD + 1 =


⇒ tan2∠CAD + 1


= cosec2∠ADC


[∠ADC = ∠BAC, from the properties]


= cosec2∠BAC


=



Question 20.

The value of (sin 43° cos47° + cos43° sin47°)
A. 0

B. 1

C. sin4°

D. cos4°
Answer:

⇒ Option B is correct as it satisfy the value


Given, sin43°cos47° + cos43°sin47°


Need to find the value


⇒ sin43°cos(90 - 43)° + cos43°sin(90 - 43)°


⇒ sin43°sin43° + cos43°cos43°


⇒ sin243° + cos243°


= 1


[Since, sin2θ + cos2θ = 1]


⇒ Option A is incorrect because by solving the equation we get the value as 1


⇒ Option C is incorrect because by solving the equation we get the value as 1


⇒ Option D is incorrect because by solving the equation we get the value as 1
Question 21.

The value of is
A. 0

B. 1

C. 2

D. none of this
Answer:

Given,


⇒


⇒


⇒


= 1 + 1


= 2


⇒ Option A is incorrect as it does not match the value 2


⇒ Option B is incorrect as it does not match the value 2


⇒ Option C is correct as it match the value 2


⇒ Option D is incorrect as it does not match the value 2
Question 22.

The value of {cos(40° + θ) - sin(50° - θ)} is
A. 2cosθ

B. 7sinθ

C. 0

D. 1
Answer:

Given, cos(40° + θ) - sin(50° - θ)


[cos(90° - θ) = sinθ ]


⇒ cos(90° - (50° - θ)) - sin(50° - θ)


⇒ sin(50° - θ) - sin(50° - θ)


= 0


⇒ Option C is the correct as the value of cos(40° + θ) - sin(50° - θ) is 0


⇒ Option A is incorrect, since it does not match the value


⇒ Option B is incorrect, since it does not match the value


⇒ Option D is incorrect, since it does not match the value
Question 23.

ABC is triangle. Sin
A. sin

B. cos

C. sinA

D. cosA
Answer:

In a triangle A+B+C =180°


(B+C)/2 = 90°-A/2


Sin cos (As sin(90°-A) = cos A)
Question 24.

If A + B = 90° and tan A = , value of cot B is
A.

B.

C.

D.


Answer:

Given, A + B = 90°


And tan A =


⇒ cotB = cot(90 - A)


= tan A


=


∴ Option A is correct, the value of cot B is also


⇒ Option B is incorrect, as it does not match the given value


⇒ Option C is incorrect, as it does not match the given value


⇒ Option D is incorrect, as it does not match the given value
Question 25.

Let us write whether the following statements are true of false:

(i) The value of cos54° and sin36° are equal.

(ii) The simplified value of (sin 12° - cos78°) is 1.
Answer:

(i) The statement is true


⇒ cos54° = cos(90 - 36)°


= sin36°


[cos(90 - θ) = sinθ]


∴ the value of cos54° and sin36° has same values


(ii) The given statement is False


⇒ sin12° - cos78°


⇒ sin12° - cos(90 - 12)°


⇒ sin12° - sin12°


= 0
Question 26.

Let us fill up the blanks:

(i) The value of (tan 15° x tan 45° x tan 60° tan 75°) is_______.

(ii) The value of (sin 12° x cos 18° x sec 78° cosec 72°) is _______.

(iii) If A and B are complementary to each other, siin A = _______.
Answer:

(i) The value of tan15° tan45° tan60° tan75° is √3


⇒ we know that tan45° = 1 and tan60° = √3


⇒ tan15°(1)(√3)tan(90 - 15)


⇒ √3 × tan15° cot15°


⇒ √3 × tan15°


= √3 × 1


= √3


(ii) The value of sin12°cos18° × sec78°cosec72° is 1


⇒ sin12°cos18° sec78°cosec72°


⇒ sin12°cos18°sec(90 - 12)°cosec(90 - 18)°


⇒ sin12°cos18°cosec12°sec18°


⇒ sin12°cos18°


= 1


(iii) sinA = cosB


⇒ If A and B are complementary angles then A + B = 90° and A = 90 - B


⇒ sinA = sin(90 - B)


= cosB
Question 27.

If sin 10θ = cos8θ and 10θ is a positive acute angle, let us find the value of tan 9θ.
Answer:

Given, sin10θ = cos8θ


And 10θ is a positive acute angle


⇒ sin10θ = cos8θ


⇒ cos(90 - 10θ) = cos8θ


⇒ 90 - 10θ = 8θ


⇒ 90 = 18θ


⇒ θ = 5


Hence, the value of θ is 5
Question 28.

If tan 4θ × tan 6θ = 1 and 6θ is a positive acute angle, let us find the value of θ.
Answer:

Given, 6θ as a positive acute angle


⇒ tan4θ tan6θ = 1


⇒ tan4θ cot(90 - 6θ) = 1


⇒


⇒ cot(90 - 6θ) = cot4θ


⇒ 90 - 6θ = 4θ


⇒ 10θ = 90


⇒ θ = 9
Question 29.

let us find the value of
Answer:

Given,


⇒


⇒


⇒


⇒


=


= 3
Question 30.

let us find the value of (tan 1° × tan2° × tan 3°………..tan89°)
Answer:

⇒ tan1° tan2° tan3° ……tan87° tan88° tan89°


⇒ tan1° tan2° tan3° ……tan(90 - 3)° tan(90 - 2)° tan(90 - 1)°


⇒ tan1° tan2° tan3° …..cot3° cot2° cot1°


= 1
Question 31.

If sec 5A = cosec (A + 36°) and 5A is a positive acute angle, let us find the value of A.
Answer:

Given, sec5A = cosec(A + 36°)


And 5A is positive acute angle


⇒ sec5A = cosec(A + 36°)


⇒ cosec(90° - 5A) = cosec(A + 36°)


[secθ = cosec(90 - θ)]


⇒ 90 - 5A = A + 36


⇒ 90 - 36 = 6A


⇒ 6A = 54


⇒ A = 9


Hence, the value of A is 9