Question

cosA/1+sinA+ 1+sinA/cosA= 2sec A prove RHS​

Answer 1

Given :

$\tt\dagger\:\frac{ \cos \: A}{1 + \sin \: A } + \frac{1 + \sin \: A}{ \cos \: A} = 2 \sec \: A.$

Solution :

Taking LHS,

$\tt \longmapsto\frac{ \cos \: A}{1 + \sin \: A } + \frac{1 + \sin \: A }{ \cos \: A}$

$\tt\longmapsto\frac{ { \cos}^{2} A + {(1 + \sin \: A)}^{2} }{(1 + \sin \: A) \cos \: A}$

$\tt\longmapsto \frac{ { \cos}^{2}A + 1 + { \sin }^{2} A + 2 \sin \: A }{(1 + \sin \: A) \cos \: A }$

$\tt\longmapsto \frac{2 + 2 \sin \: A}{(1 + \sin \: A) \cos \: A }$

$\tt\longmapsto \frac{2(1 + \sin \: A) }{(1 + \sin \: A) \cos \: A}$

$\tt\longmapsto \frac{2}{ \cos \: A}$

$\tt\longmapsto2 \sec \: A.$

Hence Proved.

$\rule{200}1$

Some Identities

• Sin²theta + Cos² theta = 1.
• 1 + tan² theta = Sec² theta.
• 1 + Cot² theta = Cosec² theta.

Answer 2

QuEsTiOn :-

Prove that

$\dfrac{cosa}{1 + sina} + \dfrac{1 + sina}{cosa} = 2seca$

Trigonometry formula Used

$\implies {sin}^{2} \theta + {cos}^{2} \theta = 1$

$\implies {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\implies \: \frac{1}{cos \theta} = sec \theta$

SoLuTiOn :-

Taking LHS

$\implies \frac{cosA}{1 + sinA} + \frac{1 + sinA}{cosA} \\ \\ By \: cross \: multiplication \\ \\ \implies \frac{ {cos}^{2}A + {(1 + sinA)}^{2} }{(1 + sinA)cosA} \\ \\ \implies \frac{ {cos}^{2}A + 1 + {sin}^{2}A + 2sinA }{(1 + sinA)cosA} \\ \\ \implies \frac{1 + 1 + 2sinA}{(1 + sinA)cosA} \\ \\ \implies \frac{2 + 2sinA}{(1 + sinA)cosA} \\ \\ \implies \frac{2( 1 + sinA)}{(1 + sinA)cosA} \\ \\ \implies \frac{2}{cosA} \\ \\ \implies2secA \: = RHS$

Here ,

${\purple{\boxed{\large{\bold{LHS = RHS}}}}}$

Hence Proved ✓