Math, asked by gauravkumar001525, 3 months ago

(tanA+cosB)² - (cotB-secA)² =2tanA.cotB (cosB+secA​

Answers

Answered by chordiasahil24
0

Step-by-step explanation:

solving for LHS

(tan²A + cos²B + 2tanAcosB) - (cot²B + sec²A - 2cotAsecB)

Lhs: tan2A+cosec2B+2 tanA.cosecB-cot2B-sec2A+2cotB. SecA

=cosec2B-cot2B-(sec2A-tan2A)+2tanA. CosecB+2cotB. SecA

=1-1+2tanA. CosecB+2cotB. SecA

=2 (sinA/cosA × 1/sinB)+2 (cosB/sinB × 1/cosA)

=2 (sinA+cosB)/cosA.sinB

Rhs:2×sinA/cosA×cosB/sinB ×{1/SinA +1/cosB}

=2( sinA. CosB/cosA.sinB)×(cosB+sinA/sinA.cosB)

=2 (sinA+cosB)/cosA. SinB

Lhs=Rhs

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