Question

Prove that:
(sin 0 + 1 + cos 0) (sin 0 - 1 + cos 6). sec 0 cosec 0 = 2​

Answer 1

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L.H.S = (1+TAN0+COT0)(SIN0-COS0) =(1+SIN/COS+COS/SIN)(SIN-COS) =

(COSSIN+SIN^2+COS^2/COSSIN)(SIN-COS) =

COSSIN^2-COS^2SIN+SIN^3-SIN^2COS+COS^2SIN-COS^3/COSSIN

=SIN^3 -COS^3/COSSIN =

R.H.S =

SEC/COSEC^2-COSEC/SEC^2 =

1/COS/1/SIN^2 -1/SIN/1/COS^2

=SIN^3-COS^3/SINCOS .

LHS=THE HENCE PROVED

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