if tanA=1/underroot7 , show that (cosec2A - sec2A)/(cosec2A = sec2A) = 3/4 - Maths -

Question

if tanA=1/underroot7 , show that
(cosec2A - sec2A)/(cosec2A =
sec2A) = 3/4

Chetna Khanna · Accepted Answer

$\tan A = \frac{1}{\sqrt{7}} ∴ \frac{1}{\sqrt{7}} = \frac{BC}{AB}$

â‡’ BC = 1 $AB = \sqrt{7} {AC}^{2} = {AB}^{2} + {BC}^{2} \Rightarrow {AC}^{2} = (\sqrt{7})^{2} + 1^{2} = 7 + 1 = 8$ $\Rightarrow AC = \sqrt{8} = 2 \sqrt{2} \sec A= \frac{AC}{AB} = \frac{2 \sqrt{2}}{\sqrt{7}} Cosec A = \frac{AC}{BC} = \frac{2 \sqrt{2}}{1}$ $LHS = \frac{(\cos^{2} A - \sec^{2} A)}{(\cos^{2} A + \sec^{2} A)} = \frac{[(2 \sqrt{5})^{2} - (\frac{2 \sqrt{2}}{\sqrt{7}})^{2}]}{[(2 \sqrt{2})^{2} + (\frac{2 \sqrt{2}}{\sqrt{7}})^{2}]}$ $= \frac{(8 - \frac{8}{7})}{(8 + \frac{8}{7})} = \frac{\frac{(56 - 8)}{7}}{\frac{(56 + 8)}{7}} = \frac{48}{64} = \frac{3}{4} = RHS$