\frac{\sin \theta-\cos\theta+1}{\sin\theta+\cos \theta-1} (where \theta\neq\pi/2 ) is equal to

Correct Answer is : 1+sinθcosθ

\frac{1+\sin\theta}{\cos\theta}

\frac{1-\sin\theta}{\cos\theta}

\frac{1-\cos\theta}{\sin\theta}

\frac{1+\cos\theta}{\sin\theta}

RRB Junior Engineer Stage 1 Model Papers 4

Question : 23 of 99

Marks: +1, -0

{\sin θ-\cosθ+1}/{\sinθ+\cos θ-1}

(where

θ≠π/2

) is equal to

Solution:

{\sin θ-\cosθ+1}/{\sinθ+\cos θ-1}

Dividing Numerator and Denominator by cos θ

={{\sinθ}/{\cosθ}-{\cosθ}/{\cosθ}+1/{\cosθ}}/{{\sinθ}/{\cosθ}+{\cosθ}/{\cosθ}-1/{\cosθ}}= {\tan θ-1+\secθ} /{\tan θ+1-\secθ}

⇒{(\tan θ+\secθ)-(\sec^2 θ-\tan^2 θ)}/{\tan θ-\secθ+1}

⇒\tan θ+\secθ

⇒{\sinθ}/{\cosθ}+1/{\cosθ}={1+\sinθ}/{\cosθ}

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