Math, asked by gurlesseleadiyakan, 1 year ago

(sin θ +cos θ )(sec θ +cosec θ )= 2+sec θ cosec θ

Answers

Answered by sangharsh1234
1
LHS = (sin A+ cos A)(secA+cosec A) sinA×secA+sinA×cosec A+cosA×secA+cosA×cosec A = tan A+1+1+cot A = 2 +( tan A+1/tanA ) =2+ ( tan ^2 A+1÷tanA) =2+sec^2A/ tanA =2+sec^2A× cosA÷sin A =2+sec A×cos A= RHS . .....PROVED
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