Q25 of 268 Page 4

Prove that :

sin 42°. cos 48° + cos 42° . sin 48° = 1

Taking LHS

= sin 42° cos 48° + cos 42° sin 48°


= cos (90° - 42°) cos 48° + sin (90° - 42°) sin 48°


[∵ Sin θ = cos (90° - θ) and cos θ = sin (90° - θ) ]


= cos 48° cos 48° + sin 48° sin 48°


= cos2 48° + sin2 48°


= 1 [∵ cos2 θ + sin2 θ = 1]


=LHS=RHS


Hence Proved


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