Question

Find the value of the following expression:
$\frac{cos{38}^{o}cosec{52}^o}{tan{18}^{o}tan{35}^{o}tan{60}^{o}tan{72}^{o}tan{55}^o}$ .

• A. $\frac{1}{tan{30}^o}$
• B. $\frac{1}{tan{60}^o}$
• C. $\frac{1}{cosec{60}^o}$
• D. $\frac{1}{cos{60}^o}$

Answer 1

The correct option is B $\frac{1}{tan{60}^o}$

Given:

$\frac{cos{38}^o\times cosec{52}^o}{tan{18}^o\times tan{35}^o\times tan{60}^o\times tan{72}^o\times tan{55}^o}$

$=\frac{cos(90-52{)}^o\times cosec{52}^o}{tan(90-72{)}^o\times tan(90-55{)}^o\times tan{60}^o\times tan{72}^o\times tan{55}^o}$

We know that, $cos(90-\theta {)}^o=sin\theta$ and $tan({90}^o-\theta )=cot\theta$.

$=\frac{sin{52}^o\times cosec{52}^o}{cot{72}^o\times cot{55}^o\times tan{60}^o\times tan{72}^o\times tan{55}^o}$


$(cosec\theta =\frac{1}{sin\theta },cot\theta =\frac{1}{tan\theta })$

$=\frac{sin{52}^o\times \frac{1}{sin{52}^o}}{\frac{1}{tan{72}^o}\times \frac{1}{tan{55}^o}\times tan{60}^o\times tan{72}^o\times tan{55}^o}$

By cancelling out the common terms in both numerator and denominator, we get,

$=\frac{1}{tan{60}^o}$

$=\frac{1}{\sqrt{3}}$ $(\because tan{60}^{\circ }=\sqrt{3})$

Hence, $\frac{cos{38}^o\times cosec{52}^o}{tan{18}^o\times tan{35}^o\times tan{60}^o\times tan{72}^o\times tan{55}^o}=\frac{1}{tan{60}^o}=\frac{1}{\sqrt{3}}$