Question

Please guys help me with this question .. ans as soon as possible ​

Answer 1

Answer:

$(\tan \theta \sin \theta+\cos \theta)^2-\left(\dfrac{sec\ \theta}{cosec\ \theta} \right)^2=1$

Step-by-step explanation:

We can simplify term by term. There are some trigonometric identities.

$\tan \theta= \dfrac{\sin \theta}{\cos \theta}\\\\\sec \theta = \dfrac{1}{\cos \theta}\\\\ cosec\ \theta = \dfrac{1}{\sin \theta}\\\\\sin^2 \theta+\cos^2 \theta= 1$

Using the above identities we can simplify the given expression. The steps are given below.

$(\tan \theta \sin \theta+\cos \theta)^2-\left(\dfrac{sec\ \theta}{cosec\ \theta} \right)^2= (\dfrac{\sin \theta}{\cos \theta}\times \sin \theta+\cos \theta)^2-\left(\dfrac{\sin\theta}{\cos \theta} \right)^2\\\\ = (\dfrac{\sin^2 \theta}{\cos \theta}+\cos \theta)^2-\left(\dfrac{\sin^2\theta}{\cos^2 \theta} \right) = (\dfrac{\sin^2 \theta+\cos^2 \theta}{\cos \theta})^2-\left(\dfrac{\sin^2\theta}{\cos^2 \theta} \right)\\\\= (\dfrac{1}{\cos \theta})^2 - \left(\dfrac{\sin^2\theta}{\cos^2 \theta} \right)$

$= (\dfrac{1}{\cos^2 \theta})(1-\sin^2 \theta) = \dfrac{\cos^2 \theta}{\cos^2\theta} =1$