Question

Evaluate: $\frac{-\tan \theta \cot ({90}^o-\theta )+\sec \theta cosec({90}^o-\theta )+{\sin }^2{35}^o+{\sin }^2{55}^o}{\tan {10}^o\tan {20}^o\tan {30}^o\tan {70}^o\tan {80}^o}$.

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

Answer 1

The correct option is B $2\sqrt{3}$

as we know $cot({90}^0$ $-$ $\theta$ $)$ $=$ $tan$ $\theta$ and $cosec({90}^0$ $-$ $\theta$ $)$ $=$ $sec$ $\theta$

Numerator $se{c}^2$ $\theta$ $-$ $ta{n}^2$ $\theta$ $=1$ and $sin{55}^0=$ $cos{35}^0$

So using the above mentioned identity the numerator outcome is $2$ and the denominator $tan{80}^0=$ $cot{10}^0$ and $tan{10}^0\ast cot{10}^0=1$ similarly for $tan{20}^{0}cot{20}^0=1$

so denominator is $tan{30}^0=1/$ $\sqrt{3}$

Hence answer is $2\ast$ $\sqrt{3}$

Hence option B is correct.