Question

Prove the following: cot² θ - tan² θ = cosec² θ - sec² θ​

Answer 1

from LHS

cot²¢ - tan²¢

we know that ,

cot²¢ = cosec²¢ - 1

tan²¢ = sec²¢ - 1

hence ,. cosec²¢ - 1 - (sec²¢ - 1 )

cosec²¢ - 1 - sec²¢ + 1

cosec²¢ - sec²¢ RHS prooved ..

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Answer 2

To Prove :-

$\sf{ {\cot }^{2} \theta - {\tan}^{2} \theta = {\cosec }^{2} \theta - {\sec}^{2} \theta }$

Solution :-

Taking LHS Firstly ,

$\sf{\implies LHS = {\cot}^{2} \theta - {\tan}^{2} \theta }$

• $\sf{\implies \cot \theta = \frac{\cos \theta}{\sin \theta} \: and \: \tan \theta = \frac{\sin \theta}{\cos \theta} }$

Putting these values in LHS

$\sf{\implies {\frac{\cos \theta }{\sin \theta } }^{2} -{\frac{\sin \theta }{\cos \theta } }^{2} }$

Adding and Substracting 1.

$\sf{\implies {\frac{\cos \theta }{\sin \theta } }^{2} +1 -{\frac{\sin \theta }{\cos \theta } }^{2} - 1 }$

Taking minus common in last fraction and 1.

$\sf{\implies ({\frac{\cos \theta }{\sin \theta } + 1 )}^{2} - ({\frac{\sin \theta }{\cos \theta } }^{2} +1) }$

$\sf{\implies \frac{ {\cos \theta}^{2} + {\sin \theta}^{2} }{ {\sin \theta }^{2} } - \frac{ {\sin \theta}^{2} + {\cos \theta }^{2} }{ {\cos \theta}^{2} }}$

• Now we know that cos² A + sin²A = 1 , So

$\sf{\implies \frac{ 1 }{ {\sin \theta }^{2} } - \frac{ 1 }{ {\cos \theta}^{2} }}$

• $\sf{\implies \sec \theta = \frac{ 1}{\cos \theta} \: and \cosec \theta = \frac{1}{\sin \theta}}$

Now putting these values we get

$\underline{\sf{\implies LHS = {\cosec}^{2} \theta - {\sec}^{2} \theta }}$

So LHS = RHS

Hence Proved.