Question

Maths question: B part​

Answer 1

tan^2A - tan^2B


sin^2A / cos^2 A - Sin^2 B / cos^2 B


Sin^2 A-Sin^2B / Cos^2A × Cos^2 B(l.c.m.)

Sin^2 A -Sin^2B/ Cos^2A × Cos^2 B = R.H.S.

•°• L.H.S. = R.H.S.

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Answer 2

$<b>HELLO !!!$

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PROVE :

tan^2 - tan^2 B = sin^2 A - sin^2B / cos^2 A cos^2B

L.H.S

$= > \tan {}^{2} a - \tan{}^{2} b \\ \\ = > \frac{ \sin {}^{2} a }{ \cos {}^{2} a} - \frac{ \sin {}^{2}b }{ \cos {}^{2} b } \\ \\ = > \frac{ \sin {}^{2}a. \cos {}^{2} b - \sin {}^{2}b \cos {}^{2} a}{ \cos {}^{2} a \: \cos {}^{2} b } \\ \\ As \: we\: Know : \\ \sin {}^{2} x + \cos {}^{2} x = 1 \\ \ \cos {}^{2} x = 1 - \ \sin {}^{2} x \\ \\ NOw, \\ \\ = > \frac{ \sin {}^{2}a(1 - \sin {}^{2} b) - \sin {}^{2}b(1 - \sin {}^{2}a )}{ \cos {}^{2}a \: \cos {}^{2} b} \\ \\ = > \frac{ \sin {}^{2} a - sin {}^{2} a. \sin {}^{2}b - \sin {}^{2} b + \sin {}^{2} a. \sin {}^{2}b }{ \cos {}^{2} a \: \cos {}^{2} b } \\ \\ = > \frac{ \sin {}^{2} a - \sin {}^{2} b}{ \cos {}^{2}a \: \cos {}^{2} b}$

SO, L.H.S = R.H.S

HENCE PROVED ✔

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