Bihar Police Constable Model Paper 5

GIVEN:
$\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\mathit{c}\mathit{o}\mathit{s}\left(90°+\mathit{\theta }\right)}$ + $\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\mathit{s}\mathit{i}\mathit{n}\left(180°+\mathit{\theta }\right)}$ + $\frac{\mathit{t}\mathit{a}\mathit{n}\left(90°+\mathit{\theta }\right)}{\mathit{c}\mathit{o}\mathit{t}\mathit{\theta }}$
FORMULA USED:
cos(90°+ θ) = - sinθ , sin(180°+ θ) = - sinθ and tan(90°+ θ) = - cotθ
CALCULATION:
$\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\mathit{c}\mathit{o}\mathit{s}\left(90°+\mathit{\theta }\right)}$ + $\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\mathit{s}\mathit{i}\mathit{n}\left(180°+\mathit{\theta }\right)}$ + $\frac{\mathit{t}\mathit{a}\mathit{n}\left(90°+\mathit{\theta }\right)}{\mathit{c}\mathit{o}\mathit{t}\mathit{\theta }}$
⇒ sinθ/(- sinθ) + sinθ/(- sinθ) + (- cotθ)/cotθ
⇒ (-1) + (-1) + (-1)
⇒ -3