Question

(1-sin a)(1-sin b)(1-sin c)= (1+sin a)(1+sin b)(1+sin c) = k


Now find the value of k.

Answer 1

heya!!

refer to given attachment please

question is from 'Trigonometric identities '.

use these formulas :-

[]Sin²a + Cos²a = 1

()COROLLARY :-

1 - Sin²a = Cos²a

1 - Cos²a = Sin²a


Thanks

Answer 2

Hey friend, Harish here.

Here is your answer:

Given that:

(1-sin a)(1-sin b)(1-sin c)= (1+sin a)(1+sin b)(1+sin c) = k

To Find:

The value of K.

Solution:

⇒ $(1-sin a)(1-sin b)(1-sin c) = k$   - (i)

⇒ $(1+sin a)(1+sin b)(1+sin c) = k$  - (ii)

Multiply (i) & (ii):

$(1-sin a)(1-sin b)(1-sin c)(1+sin a)(1+sin b)(1+sin c) = k^{2}$

⇒ $(1+sina)(1-sina)(1+sinb)(1-sinb)(1+sinc)(1-sinc) = k^{2}$

⇒ $k^{2} = (cosa)^{2}(cosb)^{2}(cosc)^{2}$

⇒ $k = \± (cos\ a)(cos\ b)(cos\ c)$
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Hope my answer is helpful to you.