Question

If $tan\theta =1$ then, find the value of $\frac{sin\theta +cos\theta }{sec\theta +cosec\theta }.$ [3 Marks]

Answer 1

We know that, $tan45°=1$
∴ $tan\theta =tan45°$
∴ $\theta =45°$
Now, $sin\theta =sin45°=\frac{1}{\sqrt{2}}cos\theta =cos45°=\frac{1}{\sqrt{2}}sec\theta =sec45°=\sqrt{2}$ [1 Mark]
$cosec\theta =cosec45°=\sqrt{2}\frac{sin\theta +cos\theta }{sec\theta +cosec\theta }=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{\sqrt{2}+\sqrt{2}}=\frac{1}{2}$ [2 Marks]