Question

tan /sec +1 + sec+1/tan = 2 cosec​

Answer 1

Step-by-step explanation:

The given equation is:

\frac{tanA}{secA-1}+\frac{tanA}{secA+1}=2cosecA

Taking the LHS of  the above equation,  

\frac{tanA}{secA-1}+\frac{tanA}{secA+1}

=\frac{tanA(secA+1)+tanA(secA-1)}{sec^{2}A -1}

=\frac{tanAsecA+tanA+tanAsecA-tanA}{sec^{2}A -1}

=\frac{2tanAsecA}{tan^{2}A}

=\frac{2secA}{tanA}

=2secAcotA

=2{\times}\frac{1}{cosA}{\times}\frac{cosA}{sinA}

=\frac{2}{sinA}

=2cosecA=RHS

Hence proved.

Answer 2

Answer:

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