Class 11 RD Sharma Solutions – Chapter 8 Transformation Formulae – Exercise 8.2 | Set 2
Question 11. If cosec A + sec A = cosec B + sec B, prove that tan A tan B =.
Solution:
We have, cosec A + sec A = cosec B + sec B
=> sec A − sec B = cosec B − cosec A
=>
=>
=>
=> tan A tan B =
=> tan A tan B =
Hence proved.
Question 12. If sin 2A = λ sin 2B, prove that.
Solution:
We are given, sin 2A = λ sin 2B
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=>
Hence proved.
Question 13. Prove that:
(i)
Solution:
We have,
L.H.S. =
=
=
=
=
=
=
=
= cot C
= R.H.S.
Hence proved.
(ii) sin (B−C) cos (A−D) + sin (C−A) cos (B−D) + sin (A−B) cos (C−D) = 0
Solution:
We have, L.H.S. = sin (B−C) cos (A−D) + sin (C−A) cos (B−D) + sin (A−B) cos (C−D)
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 14. If, prove that tan A tan B tan C tan D = −1.
Solution:
We have,
=>
=>
=>
=>
=>
=>
=>
=>. . . . (1)
Also,
=>
=>
=>
=>
=>. . . . (2)
Dividing (1) by (2), we get,
=>
=>
=>
=> tan A tan B tan C tan D = −1
Hence proved.
Question 15. If cos (α+β) sin(γ+δ) = cos (α−β) sin(γ−δ), prove that cot α cot β cot γ = cot δ.
Solution:
We have, cos (α+β) sin(γ+δ) = cos (α−β) sin(γ−δ)
=>
=>
=>
=>
=>. . . . (1)
Also,
=>
=>
=>
=>. . . . (2)
Dividing (1) by (2), we get,
=>
=>
=> cot α cot β = tan γ cot δ
=> cot α cot β cot γ = cot δ
Hence proved.
Question 16. If y sin Ø = x sin (2θ + Ø), prove that (x + y) cot (θ + Ø) = (y − x) cot θ.
Solution:
Given, y sin Ø = x sin (2θ + Ø)
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=> tan (Ø+θ) cot θ =
=>
=> (y − x) cot θ = (x + y) cot (θ + Ø)
=> (x + y) cot (θ + Ø) = (y − x) cot θ
Hence proved.
Question 17. If cos (A+B) sin (C−D) = cos (A−B) sin (C+D), prove that tan A tan B tan C + tan D = 0.
Solution:
We are given, cos (A+B) sin (C−D) = cos (A−B) sin (C+D)
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=> −tan D = tan A tan B tan C
=> tan A tan B tan C + tan D = 0
Hence proved.
Question 18. If , prove that xy + yz + zx = 0.
Solution:
We have,= k (say)
x =
y =
z =
So, L.H.S. = xy + yz + zx
=
=
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 19. If m sin θ = n sin (θ + 2a), prove that.
Solution:
We are given, m sin θ = n sin (θ + 2a)
=>
On applying Componendo and Dividendo, we get,
=>
=>
=>
=>
=>
Hence, proved.
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