Question

tan theta/sec thetha-1 + tan thetha/ sec thetha+1 = 2 cosec thetha​

Answer 1

Step-by-step explanation:

secθ−1

tanθ

+

secθ+1

tanθ

=2cosecθ

Taking the LHS of the above equation,

\frac{tan{\theta}}{sec{\theta}-1}+\frac{tan{\theta}}{sec{\theta}+1}

secθ−1

tanθ

+

secθ+1

tanθ

=\frac{tan{\theta}(sec{\theta}+1)+tan{\theta}(sec{\theta}-1)}{sec^{2}{\theta}-1}

sec

2

θ−1

tanθ(secθ+1)+tanθ(secθ−1)

=\frac{tan{\theta}sec{\theta}+tan{\theta}+tan{\theta}sec{\theta}-tan{\theta}}{sec^{2}{\theta}-1}

sec

2

θ−1

tanθsecθ+tanθ+tanθsecθ−tanθ

=\frac{2tan{\theta}sec{\theta}}{tan^{2}{\theta}}

tan

2

θ

2tanθsecθ

=\frac{2sec{\theta}}{tan{\theta}}

tanθ

2secθ

=2sec{\theta}cot{\theta}2secθcotθ

=2{\times}\frac{1}{cos{\theta}}{\times}\frac{cos{\theta}}{sin{\theta}}2×

cosθ

1

×

sinθ

cosθ

=\frac{2}{sin{\theta}}

sinθ

2

=2cosec{\theta}2cosecθ =RHS