Q13 of 113 Page 547

Prove that


Using cos(90° + θ) = - sinθ(I quadrant cosx is positive

cosec( - θ) = - cosecθ


tan(270° - θ) = tan(180° + 90° - θ) = tan(90° - θ) = cotθ


(III quadrant tanx is positive)


Similarily sin(270° + θ) = - cosθ (IV quadrant sinx is negative


cot(360° - θ) = cotθ(IV quadrant cotx is negative)





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