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Question

Prove that, $$\sqrt {\dfrac {1+\sin A}{1-\sin A}} = \sec A + \tan A$$
Or
If $$\tan\theta+\cot \theta=2$$, find the value of $$\tan^{3}\theta+7\cot^{3}\theta$$.


Solution

$$L.H.S $$
$$\displaystyle = \sqrt \frac {1+\sin A}{1-\sin A} \\$$

$$\displaystyle =\sqrt {\frac {1+\sin A}{1-\sin A} \times \frac {1+\sin A}{1+\sin A}} \\$$

$$\displaystyle = \sqrt \frac {(1+ \sin A)^2}{1-\sin^2A} \\$$

$$\displaystyle= \sqrt \frac {(1+\sin A)^2}{\cos^2A} \\$$
$$= \dfrac {1+\sin A}{\cos A} \\$$

$$= \dfrac {1+\sin A}{\cos A} \\$$

$$= \dfrac {1}{\cos A} + \dfrac {\sin A}{\cos A} \\$$

$$= \sec A + \tan A = R.H.S \rightarrow Hence \ Proved$$

Mathematics

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