Question

Prove that ,tanA+sinA/tanA-sunA=secA+1/secA-1​

Answer 1

Answer:

LHS = RHS

Step-by-step explanation:

Given that,

RHS = $\sf{\dfrac{secA \ + \ 1}{secA \ - \ 1}}$

LHS = $\sf{\dfrac{tanA \ + \ sinA}{tanA \ - \ sinA}}$

We know that

$\boxed{\sf{tanA = \dfrac{sinA}{cosA}}}$

$\boxed{\sf{sinA = \dfrac{1}{cosecA}}}$

Substituting these values in LHS

$\sf\dfrac{\dfrac{sinA}{cosA} + \dfrac{1}{cosecA}}{\dfrac{sinA}{cosA} - \dfrac{1}{cosecA}}$

On taking LCM we get,

$\sf\dfrac{\dfrac{sinA \times cosecA + cosA}{cosA \times cosecA}}{\dfrac{sinA \times cosecA - cosA}{cosA \times cosecA}}$

On Cancelling cosA × cosecA we get,

${\sf\dfrac{sinA \ \times \ cosecA + cosA}{sinA \ \times \ cosecA - cosA}}$

We know that,

$\boxed{\sf{sinA \times cosecA = 1}}$

${\sf\dfrac{1 + cosA}{1 - cosA}}$

We know that  

$\boxed{\sf{cosA = \dfrac{1}{secA}}}$

$\sf\dfrac{1 +\dfrac{1}{secA}}{1 - \dfrac{1}{secA}}$

On taking LCM we get,

$\sf\dfrac{\dfrac{secA + 1}{secA}}{\dfrac{secA - 1}{secA}}$

On Cancelling secA we get,

LHS = $\sf\dfrac{secA + 1}{secA - 1}$

RHS =  $\sf{\dfrac{secA \ + \ 1}{secA \ - \ 1}}$

∴ LHS = RHS

Hence Proved.

Answer 2

Answer:

LHS = RHS

Step-by-step explanation:

Given that,

RHS = $\sf{\dfrac{secA \ + \ 1}{secA \ - \ 1}}$

LHS = $\sf{\dfrac{tanA \ + \ sinA}{tanA \ - \ sinA}}$

We know that

$\boxed{\sf{tanA = \dfrac{sinA}{cosA}}}$

$\boxed{\sf{sinA = \dfrac{1}{cosecA}}}$

Substituting these values in LHS

$\sf\dfrac{\dfrac{sinA}{cosA} + \dfrac{1}{cosecA}}{\dfrac{sinA}{cosA} - \dfrac{1}{cosecA}}$

On taking LCM we get,

$\sf\dfrac{\dfrac{sinA \times cosecA + cosA}{cosA \times cosecA}}{\dfrac{sinA \times cosecA - cosA}{cosA \times cosecA}}$

On Cancelling cosA × cosecA we get,

${\sf\dfrac{sinA \ \times \ cosecA + cosA}{sinA \ \times \ cosecA - cosA}}$

We know that,

$\boxed{\sf{sinA \times cosecA = 1}}$

${\sf\dfrac{1 + cosA}{1 - cosA}}$

We know that  

$\boxed{\sf{cosA = \dfrac{1}{secA}}}$

$\sf\dfrac{1 +\dfrac{1}{secA}}{1 - \dfrac{1}{secA}}$

On taking LCM we get,

$\sf\dfrac{\dfrac{secA + 1}{secA}}{\dfrac{secA - 1}{secA}}$

On Cancelling secA we get,

LHS = $\sf\dfrac{secA + 1}{secA - 1}$

RHS =  $\sf{\dfrac{secA \ + \ 1}{secA \ - \ 1}}$

∴ LHS = RHS

Hence Proved.