Question

If $\text{cosec}{39}^{\circ }=x$, the value of $\frac{1}{{\text{cosec}}^2{51}^{\circ }}+{\sin }^{2}39+{\tan }^2{51}^{\circ }-\frac{1}{{\sin }^{2}51+{\sec }^2{39}^{\circ }}$, is

• A. $\sqrt{x^2-1}$
• B. $\sqrt{1-x^2}$
• C. $x^2-1$
• D. $1-x^2$

Answer 1

The correct option is C $x^2-1$

$cosec{39}^{\circ }=x\Rightarrow \sin {39}^{\circ }=\frac{1}{x}$

$\sec {51}^{\circ }=x\Rightarrow \text{cosec}{51}^{\circ }=\frac{1}{x}$

$\sin {51}^{\circ }=\sqrt{1-{\cos }^2{51}^{\circ }}$

$=\sqrt{1-\frac{1}{x^2}}=\frac{\sqrt{x^2-1}}{x}$

$\tan {51}^{\circ }=1-{\sin }^2{39}^{\circ }=1-\frac{1}{x^2}$

$\therefore \frac{1}{cose{c}^2{51}^{\circ }}+{\sin }^2{39}^{\circ }+{\tan }^2{51}^{\circ }-\frac{1}{{\sin }^2{51}^{\circ }{\sec }^2{39}^{\circ }}$

$=\frac{x^{2-1}}{x^2}+\frac{1}{x^2}+x^2-1-\frac{x^2}{x^2-1}\times \frac{x^2-1}{x^2}$

$=\frac{x^2-1+1}{x^2}+x^2-1-1$

$=1+x^2-1-1$

$=x^2-1$