Question

answer this question...................

Answer 1

HEY MATE!........
$= > tan {}^{2} x = sec {}^{2} x - 1 \\ \\ given.......tan {}^{2} a - tan {}^{2} b \\ = sec {}^{2} a - 1 - sec {}^{2}b + 1 \\ = sec {}^{2} a - sec {}^{2} b \\ we \: know \: \: ....sec \: x = \frac{1}{cos \: x} \\ = \frac{1}{cos {}^{2}a } - \frac{1}{cos {}^{2}b } \\ = \frac{cos {}^{2}b - cos {}^{2} a}{cos {}^{2} a \: .cos {}^{2}b } \\ we \: know \: ...cos {}^{2} x = 1 - sin {}^{2} x \\ \\ = \frac{1 - sin {}^{2} b- 1 + sin {}^{2}a }{(1 - sin {}^{2}a)(1 - sin {}^{2}b) } \\ = \frac{sin {}^{2}a - sin {}^{2}b }{cos {}^{2} a.cos {}^{2}b }$
HOPE IT HELPS U DEAR........


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Answer 2

Hey bro that you at spot new question nedd folower if you want folower i will follow you pr firstly mark as brainliest.

Thanks
I will follow u but i want only 1 brainliest answer plz