Question

On simplifying $\frac{se{c}^2{28}^{\circ }-co{t}^2{62}^{\circ }}{cose{c}^2{42}^{\circ }-ta{n}^2{48}^{\circ }}$, we will get

• A. 0.0
• B. 2
• C. $\frac{1}{2}$
• D. 1

Answer 1

The correct option is D 1

$\because$ $sec({90}^{\circ }-\theta )$ = $cosec\left(\theta \right)$
$\therefore$ $sec{28}^{\circ }$ = $sec({90}^{\circ }-{62}^{\circ })$ = $cosec\left({62}^{\circ }\right)$ ---- (1)

$\because$ $cosec({90}^{\circ }-\theta )$ = $sec\left(\theta \right)$

$\therefore$ $cosec{42}^{\circ }$ = $cosec({90}^{\circ }-{48}^{\circ })$ = $sec\left({48}^{\circ }\right)$ ----(2)

By Trigonometric Identity,
$cose{c}^{2}A-co{t}^{2}A=1$
and, $se{c}^{2}A-ta{n}^{2}A=1$

On substituting (1) and (2) in given equation,

$\therefore$ $\frac{se{c}^2{28}^{\circ }-co{t}^2{62}^{\circ }}{cose{c}^2{42}^{\circ }-ta{n}^2{48}^{\circ }}$

$\Rightarrow$ $\frac{cose{c}^2{62}^{\circ }-co{t}^2{62}^{\circ }}{se{c}^2{48}^{\circ }-ta{n}^2{48}^{\circ }}$

$\Rightarrow$ $\frac{1}{1}$ [from Identities]

$\therefore$ $\frac{se{c}^2{28}^{\circ }-co{t}^2{62}^{\circ }}{cose{c}^2{42}^{\circ }-ta{n}^2{48}^{\circ }}$ = 1