Question

(1-sina)(1-sinb)(1-sinc)=(1+sina)(1+sinb)(1+sinc) then show that each side is equal to +-1​

Answer 1

Answer:

Step-by-step explanation:

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