Question

If $(1+\sin A)(1+\sin B)(1+\sin C)=(1–\sin A)(1–\sin B)(1–\sin C)$ then each side is equal to

• A. $\pm \sin A\sin B\sin C$
• B. $\pm \cos A\cos B\cos C$
• C. $\pm \sin A\cos B\cos C$
• D. $\pm \cos A\sin B\sin C$

Answer 1

The correct option is B

$\pm \cos A\cos B\cos C$

Explanation for the correct option:

To find each side of equation $\mathbf{(}\mathbf{1}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{C}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{–}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{–}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{–}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{C}\mathbf{)}$:

The given equation is $(1+\sin A)(1+\sin B)(1+\sin C)=(1–\sin A)(1–\sin B)(1–\sin C)$

Multiplying by $(1+\sin A)(1+\sin B)(1+\sin C)$ on both sides,

$\begin{array}{rl}[(1+\sin A)(1+\sin B)(1+\sin C){]}^2& =(1–{\sin }^{2}A)(1–si{n}^{2}B)(1–{\sin }^{2}C)\\ [(1+\sin A)(1+\sin B)(1+\sin C){]}^2& ={\cos }^{2}A{\cos }^{2}B{\cos }^{2}C\\ (1+\sin A)(1+\sin B)(1+\sin C)& =\pm \sqrt{co{s}^{2}A{\cos }^{2}B{\cos }^{2}C}\\ & =\pm \cos A\cos B\cos C\end{array}$

Similarly, $(1–\sin A)(1–\sin B)(1–\sin C)=\pm \cos A\cos B\cos C$

Hence option ( B): $\pm \cos A\cos B\cos C$ is the correct option.