Question

$\frac{\left(1-\tan 2°cot62°\right)}{\left(\tan 152°-cot88°\right)}=$

• A. $\sqrt{3}$
• B. $-\sqrt{3}$
• C. $\sqrt{2}–1$
• D. $1–\sqrt{2}$

Answer 1

The correct option is B

$-\sqrt{3}$

Explanation for the correct option:

Step 1. Solve the given expression:

$\frac{1-\tan 2°cot62°}{\tan 152°-cot88°}$

$=\frac{1-\tan 2°\tan (90°-62°)}{\tan (180°-28°)-\tan (90°-88°)}$ $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{(}\mathbf{90}\mathbf{°}\mathbf{-}\mathbf{A}\mathbf{)}\mathbf{=}\mathit{c}\mathit{o}\mathit{t}\mathit{A}\right]$

$=\frac{1-\tan 2°\tan 28°}{-\tan 28°-\tan 2°}$ $\left[\because \mathbf{tan}\mathbf{(}\mathbf{180}\mathbf{°}\mathbf{-}\mathbf{A}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{tan}\mathit{A}\right]$

Step 2. Divide numerator and denominator by $\mathbf{1}\mathbf{-}\mathit{t}\mathit{a}\mathit{n}\mathbf{2}\mathbf{°}\mathit{t}\mathit{a}\mathit{n}\mathbf{28}\mathbf{°}\mathbf{}$:

$=-\frac{1}{\left({\displaystyle \frac{\tan 2°+\tan 28°}{1-\tan 2°\tan 28°}}\right)}$

$=-\frac{1}{\tan (2°+28°)}$ $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{A}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{B}\mathbf{}}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{A}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{B}}\right]$

$=-\frac{1}{\tan 30°}$

$\mathbf{=}\mathbf{-}\sqrt{\mathbf{3}}$

​Hence, Option ‘B’ is Correct.