Question

please solve the problem wrong answer will be reported ​

Answer 1

Answer:

I can solved this problem . to know this answer please see this following steps

Answer 2

$Given \: \theta + \alpha = 90\degree$

$\implies \alpha = 90\degree - \theta \: --(1)$

$i ) \frac{tan \theta tan\alpha + tan\theta cot \alpha }{ sin \theta sec \alpha } - \frac{sin^{2} \alpha }{cos^{2} \theta }$

$= \frac{tan \theta tan ( 90 - \theta ) + tan\theta cot (90 - \theta) }{ sin \theta sec (90 - \theta) } - \frac{sin^{2} (90- \theta) }{cos^{2} \theta }$

$= \frac{tan \theta cot \theta + tan\theta\times tan \theta}{ sin \theta cosec \theta} - \frac{cos^{2} \theta }{cos^{2} \theta }$

$= \frac{1 + tan^{2}\theta}{ sin \theta \times \frac{1}{sin \theta}} - 1$

$= 1 + tan^{2} \theta - 1$

$= tan^{2} \theta \: --(2)$

$LHS = \red{ \sqrt{ \Big(\frac{tan \theta tan\alpha + tan\theta cot \alpha }{ sin \theta sec \alpha } - \frac{sin^{2} \alpha }{cos^{2} \theta }}\Big)}\\= \sqrt{tan^{2} \theta } \: [From \: (1) ] \\\green {= tan\theta} \\= RHS$

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