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1.    Evaluate :

  (iii) \cos 48^{o}-\sin 42^{o}

Answers (1)

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\cos 48^{o}-\sin 42^{o}
The above equation can be written as ;
\cos (90^o-42^{o})-\sin 42^{o}....................(i)
It is known that \cos (90^o-\theta) = \sin \theta
Therefore, equation (i) becomes,
\sin42^{o}-\sin 42^{o} = 0

So, the answer is 0.

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