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Question

Prove that:$$1+\frac{tan^{2}A}{1+secA}=secA.$$


Solution

$$1+\dfrac{\tan^2A}{1+\sec A}=1+\dfrac{(\sec^2A-1)}{1+\sec A}$$
$$=1+\dfrac{(\sec A-1)(\sec A+1)}{(1+\sec A)}$$
$$=1+\sec A-1$$
$$=\sec A$$
$$\Rightarrow 1+\dfrac{\tan^2A}{1+\sec A}=\sec A$$.

1202151_1196767_ans_5b9872b721bb489db545b7a66be7977d.jpg

Maths

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