Question

Value of $\frac{3+\cot {80}^{\circ }\cot {20}^{\circ }}{\cot {80}^{\circ }+\cot {20}^{\circ }}$ is equal to

• A. $\cot {20}^{\circ }$
• B. $\tan {50}^{\circ }$
• C. $\cot {50}^{\circ }$
• D. $\cot \sqrt{{20}^{\circ }}$

Answer 1

The correct option is B $\tan {50}^{\circ }$

$k=\frac{3+\cot {80}^{}\cot {20}^{}}{\cot {80}^{}+\cot {20}^{}}=\frac{3+\cot ({90}^{}-{10}^{})\cot ({90}^{}-{70}^{})}{\cot ({90}^{}-{10}^{})+\cot ({90}^{}-{70}^{})}$
$\Rightarrow k=\frac{3+\tan {10}^{}\tan {70}^{}}{\tan {10}^{}+\tan {70}^{}}$
$\Rightarrow k=\frac{3\cos {10}^{}\cos {70}^{}+\sin {70}^{}\sin {10}^{}}{\sin {10}^{}\cos {70}^{}+\cos {10}^{}\sin {70}^{}}$
$\Rightarrow k=\frac{3\left(\frac{\cos ({10}^{}+{70}^{})+\cos ({10}^{}-{70}^{})}{2}\right)+\left(\frac{\cos ({10}^{}-{70}^{})-\cos ({10}^{}+{70}^{})}{2}\right)}{\sin ({10}^{}+{70}^{})}$
$\Rightarrow k=\frac{2\cos {60}^{}+\cos {80}^{}}{\sin {80}^{}}=\frac{1+\cos {80}^{}}{\sin {80}^{}}$
$\Rightarrow k=\frac{2{\cos }^2{40}^{}}{2\sin {40}^{}\cos {40}^{}}=\cot {40}^{}=\tan {50}^o$
Ans: B