Question

Q.32. Prove that-

(tanA+ cosec B)2

- (Cot B – Sec A )2

= 2 tan A cot B (Cosec A + Sec B )



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Answer 1

Explanation:

(tan a +cosec b)²-(cot b-sec a)²= 2 tan a cot b (cosec a +sec b)

= (tan A+cosec B)²- (cot B-sec A)²

= tan² A + 2 tan A cosec B+ cosec²B -[cot² B-2 cot B Sec A + Sec² A]

=[ tan²A+2tanA cosec B+ cosec² B - cot² B+ 2 cot B sec A - sec² a]

we know ( tan² A+1=sec² À)

( cot A+1= cosec²A)

= (tan²A-sec²A) +(cosec² B-cot²B) +2 tan A cosecB+2 cot B sec A

-1+2 [ tan A cosec B + cot B sec A] 2 tan A cos B

$2 \frac{tanacotb}{seca cotb } tan \: a \: cosec \: b \: + cot \: b \: sec \: a \\ = 2tan \: a \: cot \: b \frac{tan \: a \: cosec \: b \: }{tan \: a \: cot \: b } + \frac{cot \: b \: sec \: a \: }{tan \: a \: cot \: b} \\ = 2 \: tan \: a \: cot \: b \: \frac{1}{sin \: b}. \frac{sin \: b}{cos \: b} + \frac{1}{cos \: a \:} . \frac{cos \: a}{sin \: b} \\ = 2tan \: a \: cot \: |sec \: b + cosec \: a|$

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