Question

1+cos a + sin a / 1+cos a - sin a = 1+sin a/ cos a​

Answer 1

Step-by-step explanation:

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Answer 2

Given :

$:\implies\bf{\dfrac{1 + cos A + sin A}{1+cos A - sin A} = \dfrac{1 + sin A}{cos A}}$

Proof :

Taking L. H. S

$:\implies\bf{\dfrac{1 + cos A + sin A}{1+cos A - sin A}}$

Dividing Numerator and Denominator by Cos A

$:\implies\sf{\dfrac{1}{Cos A} + \dfrac {\cancel Cos A} {\cancel Cos A} + \dfrac{Sin A}{Cos A} \div \dfrac{1}{cos A} + \dfrac {\cancel Cos A} {\cancel Cos A} - \dfrac{sin A}{cos A}}$

Identity Used :

• $\implies\bf{ \dfrac{1}{Cos A} = Sec A}$
• $\implies\bf{ \dfrac{Sin A}{Cos A} = Tan A}$

$:\implies\sf {\dfrac{Sec A + 1 + tan A}{Sec A + 1 - tan A}}$

Rationalising with Denominator

That means multiply the Denominator after changing its sign.

$:\implies\sf{\dfrac{Sec A + 1 + tan A}{(Sec A + 1) - tan A} \times \dfrac{Sec A + 1 + tan A}{Sec A + 1 + tan A}}$

Identity Used :

• $\implies\bf{Sec^{2} A - Tan^{2} A =1}$
• $\implies\bf{1+ Tan^{2} A = Sec^{2}A}$

$:\implies\sf {\dfrac{(Sec A + 1 + tan A)^{2}} {(Sec A + 1)^{2} - tan^{2} A}}$

Identity Used :

• $\implies\sf{(a + b + c) ^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca}$

$:\implies\sf {\dfrac{Sec^{2} A + 1 + tan^{2} A + 2Sec A + 2Tan A + 2Sec A Tan A}{1 + Sec^{2} A + 2Sec A - Tan^{2} A}}$

$:\implies\sf {\dfrac{Sec^{2} A + Sec^{2}A + 2Sec A + 2Tan A + 2Sec A Tan A}{2 + 2Sec A}}$

$:\implies\sf {\dfrac{2Sec^{2} A + 2SecA + 2Tan A + 2Sec A Tan}{2 + 2Sec A}}$

Taking 2 common -

$:\implies\sf {\dfrac{\cancel 2[sec A (Sec A + 1) + Tan A(1 + Sec A) ]}{\cancel 2(1 + Sec A)}}$

$:\implies\sf {\dfrac{(Sec A + Tan A) (Sec A + 1)} {(1 + Sec A)}}$

$:\implies\sf {Sec A + Tan A}$

$:\implies\sf {\dfrac{1}{cos A} + \dfrac{Sin A}{Cos A}}$

Identity Used :

• $\implies\bf{Sec A = \dfrac{1}{Cos A}}$
• $\implies\bf{ Tan A = \dfrac{SinA}{Cos A}}$

Taking L. C. M

$:\implies\bf {\dfrac{1 + Sin A}{Cos A}}$

Hence,

$\boxed{\therefore{\sf {\dfrac{1 + Sin A}{Cos A} = RHS}}}$

Proved!!