Question
If tanθ = 1 then, find the values of $\frac{\sin \theta+\cos \theta}{\sec \theta+\operatorname{cosec} \theta}$
✓
Answer
Given,
tan θ = 1
⇒ θ = 45° [as tan 45° = 1]
Also,
$\frac{\sin \theta+\cos \theta}{\sec \theta+\operatorname{cosec} \theta}$
$=\frac{\sin \theta+\cos \theta}{\frac{1}{\cos \theta}+\frac{1}{\sin \theta}}$
$=\frac{\sin \theta+\cos \theta}{\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}}$
$=\sin \theta \cos \theta$
$=\sin 45^{\circ} \cos 45^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2}$
tan θ = 1
⇒ θ = 45° [as tan 45° = 1]
Also,
$\frac{\sin \theta+\cos \theta}{\sec \theta+\operatorname{cosec} \theta}$
$=\frac{\sin \theta+\cos \theta}{\frac{1}{\cos \theta}+\frac{1}{\sin \theta}}$
$=\frac{\sin \theta+\cos \theta}{\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}}$
$=\sin \theta \cos \theta$
$=\sin 45^{\circ} \cos 45^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2}$
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