Question

(1+tanA+secA)(1+cotA-cosecA)=2​

Answer 1

Given: (1+tanA+secA)(1+cotA-cosecA) = 2​

To find: Whether LHS = 2

Solution: LHS can be written as:

⇒ (1 + sinA/cosA + 1/cosA)(1 + cosA/sinA - 1/sinA)  

⇒ [(cosA + sinA + 1)/cosA] [(sinA+cosA - 1)/sinA]    

⇒ [$(sinA + cosA)^{2}$ - 1]/sinAcosA    

⇒ (1 + 2sinAcosA - 1) / sinAcosA  

⇒ 2

LHS = RHS

Answer 2

Step-by-step explanation:

We have,

$\rm (1+ \tan A + \sec A )(1+ \cot A -\cosec A)$

$\rm \bigg( 1+ \dfrac{\sin A}{\cos A} + \dfrac{1}{\cos A} \bigg) \bigg( 1+ \dfrac{\cos A}{\sin A} -\dfrac{1}{\sin A} \bigg)$

$\rm \bigg( \dfrac{\cos A + \sin A+1}{\cos A} \bigg) \bigg( \dfrac{\sin A + \cos A- 1}{\sin A} \bigg)$

$\rm \bigg( \dfrac{( \cos A + \sin A )^2 -1^2)}{\cos A \sin A} \bigg)$

$\rm \bigg( \dfrac{\cos ^2 A + \sin ^2 A + 2 \sin A \cos A -1}{\cos A \sin A} \bigg)$

$\rm \bigg( \dfrac{1 + 2 \sin A \cos A -1}{\cos A \sin A} \bigg)$

$\rm \dfrac{2 \: \: \cancel{\sin A \cos A} }{\cancel{\cos A \sin A}} = 2$

$\bf Hence \: Proved$