Math, asked by tejal77, 11 months ago

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

Answers

Answered by Tomboyish44
7

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.


tejal77: thank you so much
Tomboyish44: No Problem! If you have any doubts, feel free to ask me :)
Anonymous: I think some latex error !! ?
Tomboyish44: Which part?
Tomboyish44: Yeah, I see it through Mobile.
tejal77: what ?
Answered by EmperorSoul
0

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.

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