Question

sin theta + tan theta upon cos theta is equals to tan theta in bracket 1 + sec theta plss answer this​

Answer 1

$\large \dag$ Question :-

Prove that ;

$\boxed{ \boxed{ \rm \frac{sin \theta + tan\theta}{cos\theta} = tan\theta(1 + sec\theta) }} \bigstar$

$\large \dag$ Step by step Solution :-

Taking Right Hand Side (RHS)

$\: \: \: \: \rm tan\theta(1 + sec\theta)$

$\small \rm = tan\theta \bigg(1 + \frac{1}{cos\theta} \bigg) \: \: \bigg\{\red{\because \sf sec\theta = \frac{1}{cos\theta} } \bigg\}$

$\rm = tan\theta + \frac{tan\theta}{cos\theta}$

$\small \rm = \frac{sin\theta}{cos\theta} + \frac{tan\theta}{cos\theta} \: \: \bigg\{\red{\because \sf tan\theta = \frac{ sin\theta}{cos\theta} } \bigg\}$

⏩ Taking cosθ as LCM ;

$\\ \large \pmb{ \purple{\rm = \frac{sin\theta + tan\theta}{cos\theta} } }$

which is your Left Hand Side (LHS)

$\large \pink \maltese \: \: \underline{\orange{\underline{\frak{\pmb{\text Hence\:\:Proved }}}}}$

Answer 2

Correct Question :-

Prove that

$\pmb{\boxed{ \rm \frac{sin \theta + tan\theta}{cos\theta} = tan\theta(1 + sec\theta) }}\LARGE \dag$

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Solution with explanation :-

LHS 》

$\frac{sin \theta + tan\theta}{cos\theta} \\ \\ = \frac{sin \theta}{cos\theta} + \frac{tan\theta }{{cos\theta} } \\ \\ = \frac{sin \theta}{cos\theta} + tan\theta \times \frac{1 }{{cos\theta} } \\ \\ = tan\theta + tan\theta \times sec \theta \\ \\ \pink{ \large \pmb = \pmb{ tan\theta(1 + sec\theta)}} \\ \\ \red{ \rm \Large = RHS}$

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Used Formulae :-

$\rm \bigstar \: \frac{sin\theta}{cos\theta} = tan\theta \\ \\ \rm\bigstar \: \frac{1}{cos\theta} = sec\theta$