Question

please answer correct and fast​

Answer 1

According to the question,

$\begin{array}{l} \displaystyle \rm \pmb{ L.H.S. }=\frac{1}{(\sec \theta-\tan \theta)}-\frac{1}{\cos \theta} \\ \\ \displaystyle \rm=\frac{1}{(\sec \theta-\tan \theta)} \times \frac{(\sec \theta+\tan \theta)}{(\sec \theta+\tan \theta)}-\sec \theta \\ \\ \displaystyle \rm =\frac{(\sec \theta+\tan \theta)}{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}-\sec \theta \\ \\ \displaystyle \rm =(\sec \theta+\tan \theta)-\sec \displaystyle \rm \theta \qquad \left[\therefore \sec ^{2} \theta-\tan ^{2} \theta=1\right] \\ \\ \displaystyle \rm =\tan \theta \end{array}$

$\\ \pmb{ \rm R.H.S. }\rm =\dfrac{1}{\cos \theta}-\dfrac{1}{(\sec \theta+\tan \theta)}$

$\[ \begin{array}{l} \displaystyle\rm =\sec \theta-\frac{1}{(\sec \theta+\tan \theta)} \times \frac{(\sec \theta-\tan \theta)}{(\sec \theta-\tan \theta)} \\ \\ \\ \\ \displaystyle\rm=\sec \theta-\frac{(\sec \theta-\tan \theta)}{\left(\sec ^{2} \theta-\tan ^{2} \theta\right)} \\ \\ \\ \\ \displaystyle\rm=\sec \theta-(\sec \theta-\tan \theta) \quad[\because \left.\sec ^{2} \theta-\tan ^{2} \theta=1\right] \\ \\ \\ \\ \displaystyle\rm=\tan \theta . \\ \\ \\ \\ \displaystyle\rm\therefore \text { L. H.S. }=\text { R.H.S. } \end{array} \]$

Answer 2

Answer:

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