(Sin a+cos a)(tan a+cot a)= sec a+ cosec a

1.	(Sin a+cos a)(tan a+cot a)= sec a+ cosec a
Answer» Given: $\bf (sin \: a + cos \: a)(tan \: a + cot \: a) = sec \: a + cosec \: a$ To PROVE : RHS = LHS Proof : Let's take LHS :- $\sf \implies(sin \: a + cos \: a)(tan \: a + cot \: a)$ We KNOW that :- $\underline{ \boxed{ \bf \large \: tan \: x = \dfrac{sin \: x }{cos \: x}}}$ $\underline{ \boxed{ \bf \large \:co t \: x = \dfrac{cos\: x }{sin \: x}}}$ $\sf \implies(sin \: a + cos \: a) \bigg( \dfrac{sin \: a }{cos \: a} + \dfrac{ cos\: a }{ sin\: a} \bigg)$ $\sf \implies(sin \: a + cos \: a) \bigg( \dfrac{ {sin}^{2} a+ {cos}^{2}a }{ cos\: a \: sin\: a} \bigg)$ We know that :- $\underline{ \boxed{ \bf \large {sin}^{2} x+ {cos}^{2}x = 1}}$ $\sf \implies(sin \: a + cos \: a) \bigg( \dfrac{ 1 }{ cos\: a \: sin\: a} \bigg)$ $\sf \implies \bigg( \dfrac{ sin \: a}{ cos\: a \: sin\: a} \bigg) + \bigg( \dfrac{ cos\: a}{ cos\: a \: sin\: a} \bigg)$ $\sf \implies \bigg( \dfrac{ \cancel {sin \: a}}{ cos\: a \: \: \cancel {sin \: a} } \bigg) + \bigg( \dfrac{ \cancel {cos\: a}}{ \cancel {cos\: a} \: \: sin\: a} \bigg)$ $\sf \implies \bigg( \dfrac{1}{ cos\: a } \bigg) + \bigg( \dfrac{ 1}{ sin\: a} \bigg)$ We know that :- $\underline{ \boxed{ \bf \large \dfrac{1 }{cos \: x} = sec \: x} } \\ \underline{ \boxed{ \bf \large \dfrac{1 }{sin \: x} = cosec \: x} }$ $\sf \implies sec \: a + cosec \: a$ Therefore LHS = RHS HENCE proved

Answer»

Given:

$\bf (sin \: a + cos \: a)(tan \: a + cot \: a) = sec \: a + cosec \: a$

To PROVE :

RHS = LHS

Proof :

Let's take LHS :-

$\sf \implies(sin \: a + cos \: a)(tan \: a + cot \: a)$

We KNOW that :-

$\underline{ \boxed{ \bf \large \: tan \: x = \dfrac{sin \: x }{cos \: x}}}$

$\underline{ \boxed{ \bf \large \:co t \: x = \dfrac{cos\: x }{sin \: x}}}$

$\sf \implies(sin \: a + cos \: a) \bigg( \dfrac{sin \: a }{cos \: a} + \dfrac{ cos\: a }{ sin\: a} \bigg)$

$\sf \implies(sin \: a + cos \: a) \bigg( \dfrac{ {sin}^{2} a+ {cos}^{2}a }{ cos\: a \: sin\: a} \bigg)$

We know that :-

$\underline{ \boxed{ \bf \large {sin}^{2} x+ {cos}^{2}x = 1}}$

$\sf \implies(sin \: a + cos \: a) \bigg( \dfrac{ 1 }{ cos\: a \: sin\: a} \bigg)$

$\sf \implies \bigg( \dfrac{ sin \: a}{ cos\: a \: sin\: a} \bigg) + \bigg( \dfrac{ cos\: a}{ cos\: a \: sin\: a} \bigg)$

$\sf \implies \bigg( \dfrac{ \cancel {sin \: a}}{ cos\: a \: \: \cancel {sin \: a} } \bigg) + \bigg( \dfrac{ \cancel {cos\: a}}{ \cancel {cos\: a} \: \: sin\: a} \bigg)$