Question

If $\tan {\theta }_1=k\cot {\theta }_2$, then $\frac{\cos ({\theta }_1-{\theta }_2)}{\cos ({\theta }_1+{\theta }_2)}=$

• A. $\frac{1+k}{1-k}$
• B. $\frac{1-k}{1+k}$
• C. $\frac{k+1}{k-1}$
• D. $\frac{k-1}{k+1}$

Answer 1

The correct option is A $\frac{1+k}{1-k}$

$\tan {\theta }_1=k\cot {\theta }_2$

$\Rightarrow \tan {\theta }_1=\frac{k}{\tan {\theta }_2}$

$\Rightarrow k=\tan {\theta }_1.\tan {\theta }_2$

$\Rightarrow \frac{\cos ({\theta }_1-{\theta }_2)}{\cos ({\theta }_1+{\theta }_2)}=\frac{\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2}{\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2}$

$=\frac{1+\tan {\theta }_1\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}$

$=\frac{1+k}{1-k}$

Hence, the answer is $\frac{1+k}{1-k}.$