Question

$8) question$
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Answer 1

Step-by-step explanation :

$[sec^2 \theta -1=tan^2 \theta]$

$= \frac{tan \theta}{sec \theta+1} \\\\ =\frac{tan \theta}{sec \theta+1} \times \frac{sec\theta-1}{sec \theta-1} \\\\ =\frac{tan \theta(sec \theta-1)}{(sec \theta+1)(sec\theta-1)} \\\\ = \frac{tan \theta(sec\theta-1)}{sec^2 \theta-1^2} \\\\ = \frac{tan \theta(sec\theta-1)}{sec^2 \theta-1} \\\\ = \frac{tan \theta(sec\theta-1)}{tan^2\theta} \\\\ =\frac{sec \theta-1}{tan \theta}$

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FORMULA :

(a+b)(a-b) = a² - b²