Question

If (1-sinA) (1-sinB) (1-sinC) = (1+sinA) (1+sinB) (1+sinC) then prove that the value of each = +cosA cosB cos C

Answer 1

Trigonometry

We have
(1-sinA) (1-sinB) (1-sinC) = (1+sinA) (1+sinB) (1+sinC)

Have to prove that both sides are equal to ±cosA cosB cosC (I know this problem there is a mistake is will be plus minus instead of just plus sign)

Let
(1-sinA) (1-sinB) (1-sinC) = (1+sinA) (1+sinB) (1+sinC) = k

So k²
= k × k
= (1-sinA) (1-sinB) (1-sinC) × (1+sinA) (1+sinB) (1+sinC)
= (1-sin²A) (1-sin²B) (1-sin²C)
= cos²A cos²B cos²C

So k = √(cos²A cos²B cos²C)
or, k = ± cosA cosB cosC

That's it
Hope it helped

Answer 2

hope it will help you... please mark as brainliest