Question

if tan A = a tan B and sin A = b sin B prove that cos^2 A = b^2 - 1/ a^2 -1

Answer 1

》tanA=a(tanB)
》a=tanA/tanB

》a^2=tan^2A/tan^2B .......(i)

And;

》b=sinA/sinB
》b^2=sin^2A/sin^2B ......(ii)


RHS=b^2-1/a^2-1
=sin^2A-sin^2B/sin^2B×tan^2A-tan^2B/tan^2B
=sin^2A-sin^2B/sin^2B×(sin^2A×cos^2B- sin^2B×cos^2A)/(cos^2A×cos^2B)×(cos^2B /sin^2B)
= cos^2A

HENCE PROVED

Answer 2

Answer:

Proof of the below relation is explained below

$cos {}^{2} A = \frac{b {}^{2} - 1 }{a {}^{2} - 1}$

Step-by-step explanation:

Given trigonometric relations are

$tan A = a tan B \\ sin A = b sin B$

From these two relations,we have

$\frac{tanA}{tanB} = a = > \frac{sinA}{sinB} = a \frac{cosA}{cosB} \\ \\ \frac{sinA}{sinB} = b = > \frac{cosA}{cosB} = \frac{b}{a}$

Using trigonometric identity we can write that

$sin {}^{2} B + cos {}^{2} B = 1$

From our given relations we have

$sinB = \frac{sinA}{b} \\ cosB = \frac{a}{b} cosA$

Substituting these in the above mentioned identity we get,

$( \frac{sinA}{b} ) {}^{2} + ( \frac{a}{b}cos A) {}^{2} = 1 \\ = > sin {}^{2} A + a {}^{2} cos {}^{2} A = b {}^{2} \\ = > (1 - cos {}^{2} A ) + a {}^{2} cos {}^{2} A = b {}^{2} \\ = > cos {}^{2} A (a {}^{2} - 1) = b {}^{2} - 1 \\ = > cos {}^{2} A = \frac{b {}^{2} - 1 }{a {}^{2} - 1}$

Therefore,it is proved that

$cos {}^{2} A = \frac{b {}^{2} - 1 }{a {}^{2} - 1}$

#SPJ2