Class-X Maths Introduction to Trigonometry Solved Problems
1. Q. Prove that: sin6ө + cos6ө=1- 3sin2өcos2ө
sin2ө)3 + (cos2ө)3 = (sin2ө + cos2ө)3 − 3 (sin2ө cos2ө)(sin2ө + cos2ө) [since a + b = (a+b)3 − 3ab(a+b)]
= 1 − 3sin2өcos2ө [Since sin2ө + cos2ө = 1]
2. Q. If sin A+cosA=x , prove that sin6 A +cos6 A= [4 - 3(x2 - 1)]/4
sinA + cosA = x
Squaring on both sides, we get sin2A + cos2A + 2 sin A cos A = x2
Sin A cos A = (x2-1)/2 ----(i)
sin2ө)3 + (cos2ө)3 = (sin2ө + cos2ө)3 − 3 (sin2ө cos2ө)(sin2ө + cos2ө) [since a3 + b3 = (a+b)3 − 3ab(a+b)]
= 1 − 3sin2өcos2ө [Since sin2ө + cos2ө = 1]
= 1 – 3 {( x2 – 1)/2} 2
= 1 – [3( x2 – 1) 2]/4
= [ 4 – 3( x2 – 1) 2]/4
3. Q. if cosec A - sin A=a 3 , sec A - cos A=b 3 ,prove that a2b2(a2+b2) = 1
a3 = cos2A /sinA
sinA = cos2 A /a3
b 3=sin 2A/cosA = cos4A/a 6 x 1/ cos A
so cos3A = b3 a6
 Cos A =b.a 2
similarly ....... Sin A = b2 a
so sin2 A +cos2 A = 1
4
2
2
=b a +b a
4
2
2
2
2
> a b (a + b ) = 1
4. If sec A+ tan A = p, then find the value of sec A – tan A
Given secA + tan A = p
Recall the identity, sec2A – tan2A = 1
⇒ (sec A + tan A)(sec A – tan A) = 1
⇒ p × (sec A – tan A) = 1
∴ (sec A – tan A) = 1/p
5. f (sec A + tan A)(sec B + tan B)(sec C + tan C) = (secA - tanA)(secB - tanB)(secC - tanC), prove that each of the
sides is equal to 1 or -1.
Solution: (secA + tanA)(secB + tanB)(secC + tan C) = (secA - tanA)(secB - tanB)(secC - tanC)
Mulitply both sides with "(secA + tanA)(secB + tanB)(secC + tan C)",
we get (secA + tanA)2(secB + tanB)2(secC + tan C)2
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= (sec2A - tan2A)(sec2B - tan2B)(sec2C - tan2C) = (1)(1)(1) = 1
⇒ [(secA + tanA)(secB + tanB)(secC + tan C)]2 = 1
∴ (secA + tanA)(secB + tanB)(secC + tan C) = ± 1
Similarly, we get (secA – tanA)(secB – tanB)(secC – tan C) = ± 1
6. Prove (1+cotA-cosecA)(1+tanA+secA) = 2
Solution: (1+cotA-cosecA)(1+tanA+secA)
[Since sec2 A − tan2 A = 1]
7. If sinA +sin2A+sin3A=1, prove that cos6A-4cos4A+8cos2A= 4.
Solution:
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8. Show that (1-sinA+cosA) 2=2(1+cosA)(1-sinA)
Solution: (1-sinA+cosA)2 = [(1-sinA) + cosA]2
= (1-sinA)2 + cos2A + 2(1-sinA)cosA
= 1 + (sin2A + cos2A) − 2sinA + 2(1-sinA)cosA
= 2 − 2sinA + 2(1-sinA)cosA
= 1 + sin2A − 2sinA + cos2A + 2(1-sinA)cosA
= 1 + 1 − 2sinA + 2(1-sinA)cosA [Since, sin2A + cos2A =1]
= 2(1 − sinA) + 2(1-sinA)cosA
= 2(1 − sinA)(1 + cosA)
9. Prove that sinθ - cosθ + 1/ sinθ + cosθ - 1 = 1/ secθ - tanθ, using the identity sec2 θ = 1 + tan2 θ
10. write all other trignometrical identities in form of cot A?
Consider, cosec2A − cot2A = 1  cosec2A = 1 + cot2A 
We know that
We know that
Class-X Maths Introduction to Trigonometry Solved Problems by Jsnil
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