Question

Question 9
If $tan\theta +sec\theta =l$ then prove that $sec\theta =\frac{l^2+1}{2l}$.

Answer 1

Given, $tan\theta +sec\theta =l$ ...(i)

[multiply by $(sec\theta -tan\theta )$ on numerator and denominator of the LHS]

$\Rightarrow \frac{(tan\theta +sec\theta )(sec\theta -tan\theta )}{(sec\theta -tan\theta )}=l\Rightarrow \frac{(se{c}^2\theta -ta{n}^2\theta )}{(sec\theta -tan\theta )}=l$

$\Rightarrow \frac{1}{sec\theta -tan\theta }=l[\because se{c}^2\theta -ta{n}^2\theta =1]$

$\Rightarrow sec\theta -tan\theta =\frac{1}{l}\dots \dots \left(ii\right)$

On adding Eq. (i) and (ii), we get

$2sec\theta =l+\frac{1}{l}$

$\Rightarrow sec\theta =\frac{l^2+1}{2l}$