prove that. cos(π+x)×cos(-x)÷sin (π-x)×cos(π/2+x)=cot^2x
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Answer:
LHS.
Cos(π+x)×cos(-x) ÷ sin(π-x)×cos(π/2+x)
= -cosx.cosx ÷ sinx.-sinx
= -cos^2 x ÷ -sin^2 x
= cos^2 x ÷ sin^2 x
= (cosx ÷ sinx)^2
= cot^2 x = RHS
Explanation:
cos(π+x) = -cosx
cos(-x) = cosx
sin(π-x) = sinx
cos(π/2+x) = -sinx
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