Question

( sec θ - cos θ ) ( cot θ + tan θ ) = tan θ sec θ​

Answer 1

Answer:

$\huge\mathcal\pink{ANSWER}$

sinθ(1+tanθ)+cosθ(1+cotθ)

=sinθ+ cosθsin 2 θ +cosθ+ sinθcos 2 θ

=sinθ+ sinθcos 2 θ + cosθsin 2 θ +cosθ

A= sinθsin 2 θ+cos 2 θ + cosθsin 2 θ+cos 2 θ

As,sin 2 θ+cos 2 θ=1

A= sinθ1 + cosθ1

= cosecθ+secθ

Answer 2

Required Answer:-

Given to prove:

• ( sec θ - cos θ ) ( cot θ + tan θ ) = tan θ sec θ

Proof:

Taking LHS,

$\rm ( \sec\theta - \cos \theta)( \cot \theta + \tan \theta)$

$\rm = \bigg( \dfrac{1}{ \cos \theta} - \cos \theta \bigg) \bigg( \dfrac{ \cos \theta }{ \sin \theta } + \dfrac{ \sin \theta}{ \cos \theta} \bigg)$

$\rm = \bigg( \dfrac{1 - \cos ^{2} \theta }{ \cos \theta} \bigg) \bigg( \dfrac{ \cos^{2} \theta + { \sin }^{2} \theta}{ \sin \theta \cos \theta } \bigg)$

$\rm = \dfrac{ \sin^{2} \theta }{ \cos \theta} \times \dfrac{1}{ \sin \theta \cos \theta }$

$\rm = \dfrac{ \sin \theta }{ \cos \theta} \times \dfrac{1}{ \cos \theta }$

$\rm = \tan \theta\times \dfrac{1}{ \cos \theta }$

$\rm = \tan \theta \sec \theta$

= RHS (Hence Proved)

Formula Used:

• sin(x)/cos(x) = tan(x)
• cos(x)/sin(x) = cot(x)
• sin²(x) + cos²(x) = 1
• cos(x) = 1/sec(x)
• sec(x) = 1/cos(x)