Question

If tan $\theta$ = $\frac{1}{\sqrt{7}}$ then $\frac{(cose{c}^2\theta -se{c}^2\theta )}{(cose{c}^2\theta +se{c}^2\theta )}$ = ?

(a) $\frac{-2}{3}$ (b) $\frac{-3}{4}$ (c) $\frac{2}{3}$ (d) $\frac{2}{4}$

Answer 1

tan $\theta$ = $\frac{1}{\sqrt{7}}$ then $\frac{(cose{c}^2\theta -se{c}^2\theta )}{(cose{c}^2\theta +se{c}^2\theta )}$ = ?

$tan\theta =\frac{Oppositeside}{Adjacentside}$

Let ‘x’ be the hypotenuse
By applying Pythagoras

$A{C}^2=A{B}^2+B{C}^2{x}^2=1^2+(\sqrt{7}{)}^2{x}^2=1+7=8x=\sqrt{8}=2\sqrt{2}cosec\theta =\frac{AC}{AB}=2\sqrt{2}sec\theta =\frac{AC}{BC}=\frac{2\sqrt{2}}{\sqrt{7}}$

Substitute, cosec, sec in equation

$\frac{(cose{c}^2\theta -se{c}^2\theta )}{(cose{c}^2\theta +se{c}^2\theta )}=\frac{(2\sqrt{2}{)}^2-(\frac{2\sqrt{2}}{\sqrt{7}}{)}^2}{(2\sqrt{2}{)}^2+(\frac{2\sqrt{2}}{\sqrt{7}}{)}^2}=\frac{8-4\times \frac{2}{7}}{8+4\times \frac{2}{7}}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}=\frac{48}{64}=\frac{3}{4}$