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If : (1-sin alpha)(1-sin beta)(1-singamm...

If : `(1-sin alpha)(1-sin beta)(1-singamma)=(1+sinalpha)(1+sin beta)(1+sin gamma),`
then one value of each side is

A

`cos alpha* cos beta* cos gamma`

B

`sin alpha* sin beta* sin gamma`

C

`sec alpha* sec beta* sec gamma`

D

`tan alpha* tan beta* tan gamma`

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The correct Answer is:
To solve the equation \((1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma) = (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma)\), we will manipulate both sides step by step. ### Step 1: Rewrite the equation We start with the given equation: \[ (1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma) = (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) \] ### Step 2: Use the identity \(1 - \sin^2 x = \cos^2 x\) We can rewrite the left-hand side (LHS) using the identity \(1 - \sin^2 x = \cos^2 x\): \[ (1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma) = \frac{(1 - \sin \alpha)(1 - \sin \beta)(1 - \sin \gamma)(1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma)}{(1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma)} \] ### Step 3: Simplify the LHS Now we can express the LHS as: \[ (1 - \sin^2 \alpha)(1 - \sin^2 \beta)(1 - \sin^2 \gamma) = \cos^2 \alpha \cos^2 \beta \cos^2 \gamma \] ### Step 4: Rewrite the RHS The right-hand side (RHS) can be expressed similarly: \[ (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) \] ### Step 5: Set the two sides equal Now we have: \[ \frac{\cos^2 \alpha \cos^2 \beta \cos^2 \gamma}{(1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma)} = (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) \] ### Step 6: Cross-multiply Cross-multiplying gives us: \[ \cos^2 \alpha \cos^2 \beta \cos^2 \gamma = (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma)(1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) \] ### Step 7: Final expression Thus, we find that: \[ \cos^2 \alpha \cos^2 \beta \cos^2 \gamma = (1 + \sin \alpha)(1 + \sin \beta)(1 + \sin \gamma) \] ### Conclusion From the above steps, we conclude that the equation holds true under the condition that: \[ \cos \alpha \cos \beta \cos \gamma = 1 + \sin \alpha + \sin \beta + \sin \gamma \]
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