Prove the following

Question:

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

Solution:

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

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

$\Rightarrow \quad \frac{(\tan \theta+\sec \theta)(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)}=l \quad \Rightarrow \frac{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{(\sec \theta-\tan \theta)}=l$

$\Rightarrow$ $\frac{1}{\sec \theta-\tan \theta}=l$ $\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$

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

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

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

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

Hence proved.

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