Question

Prove that tanA/secA-1 + tanA/secA+1 = 2cosecA​

Answer 1

LHS = tanA/secA - 1 + tanA/secA + 1

= sinA/cosA / 1/cosA - 1 + sinA/cosA / 1/cosA + 1 [because tan A = sin A / cos A and sec A = 1 / cos A]

= sinA/cosA / (1 - cosA) / cosA + sinA/cosA / (1 + cosA) / cos A

= sinA / 1 - cosA + sinA / 1 + cos A

= [sinA (1 + cosA) + sinA (1 - cosA)] / (1 - cosA) (1 + cosA)

= [sinA + sinAcosA + sinA - sinAcosA] / 1 - cos(sq) A

= 2sinA / sin(sq) A [From identity: sin(sq) A + cos(sq) A = 1]

= 2 / sin A

= 2 x 1 / sinA

=2 cosecA

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Answer 2

$\bf\huge\blue{\underline{\underline{ Question : }}}$

Prove that :

$\sf\: \dfrac{\tan\: A}{\sec\:A - 1} + \dfrac{\tan\:A}{\sec\:A + 1} = 2 \csc \: A$

$\bf\huge\blue{\underline{\underline{ Solution : }}}$

★ LHS :

$\sf\:\implies \dfrac{\tan\:A}{\sec\:A - 1} + \dfrac{\tan\:A}{\sec\:A + 1}$

• tan θ = sin θ/cos θ
• sec θ = 1/cos θ

$\sf\:\implies \dfrac{\cfrac{\sin\:A}{\cos\:A}}{\cfrac{1}{\cos\:A} - 1} + \dfrac{\cfrac{\sin\:A}{\cos\:A}}{\cfrac{1}{\cos\:A} + 1}$

$\sf\:\implies \dfrac{\cfrac{\sin\:A}{\cos\:A}}{\cfrac{1 - \cos\:A}{\cos\:A}} + \dfrac{\cfrac{\sin\:A}{\cos\:A}}{\cfrac{1 + \cos\:A}{\cos\:A}}$

$\sf\:\implies \dfrac{\sin\:A}{\cos\:A} \times \dfrac{\cos\:A}{1 - \cos\:A}\ + \dfrac{\sin\:A}{\cos\:A}\times \dfrac{\cos\:A}{1 + \cos\:A}$

$\sf\:\implies \dfrac{\sin\:A}{1 -\cos\:A} + \dfrac{\sin\:A}{1 + \cos\:A}$

$\sf\:\implies \dfrac{\sin\:A( 1 + \cos\:A) + \sin\:A(1 - \cos\:A)}{(1 - \cos\:A)(1 + \cos\:A)}$

$\sf\:\implies \dfrac{\sin\:A + \sin\:A\cos\:A + \sin\:A - \sin\:A\cos\:A}{(1)^{2} - (\cos\:A)^{2}}$

• (a + b)(a - b) = a² - b²

$\sf\:\implies \dfrac{\sin\:A + \sin\:A}{1 - \cos^{2}\:A}$

• 1 - cos² θ = sin² θ

$\sf\:\implies \dfrac{2 \sin\:A}{\sin^{2}\:A}$

$\sf\:\implies \dfrac{2}{\sin\:A}$

$\sf\:\implies 2 \: \dfrac{1}{\sin\:A}$

• 1/sin θ = cosec θ

$\sf\:\implies 2\:{\csc}\:A = \tt\green{RHS}$

# Hence,it was proved.

★ More information :

$\boxed{\begin{minipage}{7 cm} Fundamental Trigonometric Identities \\ \\ :\implies \sin^{2}\theta + cos^{2}\theta = 1 \\ \\ :\implies \ 1 + tan^{2}\theta = sec^{2}\theta \\ \\ :\implies 1 + cot^{2}\theta=\text{cosec}^2\, \theta \end{minipage}}$

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