Question

if secA=17/8 show that 3-4sin^2A/4cos^2A-3=3-tan^2A/1-3tan^2

Answer 1

Sec A = 17/8      =>   cosA = 8/17       => sinA = √[17² - 8²] / 17 = 15/17
Tan A = sinA/cosA = 15/8

LHS  =  [3 - 4 Sin² A ] / [4 cos² A - 3 ]
         =  [3 - 4* 225/289 ] / [4 * 64/289  - 3 ]
         =  -33 / (- 611)

RHS = [ 3 - 225/64 ] / [1 - 3 * 225/64 ]
         = [ - 33 ] / [ - 611 ] 

Hence proved.

Answer 2

Answer:

$Sec \theta = \frac{Hypotenuse}{Base}$

We are given that sec A = 17/8

On comparing

Hypotenuse = 17

Base = 8

$Hypotenuse^2= Perpendicular^2+Base^2$

$17^2= Perpendicular^2+8^2$

$\sqrt{17^2-8^2}= Perpendicular$

$15= Perpendicular$

$Sin A = \frac{Perpendicular}{Hypotenuse}=\frac{15}{17}$

$Cos A =\frac{Base}{Hypotenuse}=\frac{8}{17}$

$Tan A = \frac{Perpendicular}{Base}=\frac{15}{8}$

We are supposed to find $\frac{3-4sin^2A}{4cos^2A-3}=\frac{3-tan^2A}{1-3tan^2}$

LHS = $\frac{3-4sin^2A}{4cos^2A-3}$

LHS = $\frac{3-4(\frac{15}{17})^2}{4(\frac{8}{17})^2-3}$

LHS = $\frac{33}{611}$

RHS = $\frac{3-tan^2A}{1-3tan^2}$

RHS = $\frac{3-(\frac{15}{8})^2}{1-3(\frac{15}{8})^2}$

RHS = $\frac{33}{611}$

LHS = RHS

Hence verified