Prove that 1 - tan2-1 - 2- - 1 - tan-1 - -2

1+tan^2 θ1+^2 θ=(1 tanθ1 θ)^2 L.H.S 1+ tan^2 θ1+ ^2 θ= 1+sin^2 θcos^2 θ1+cos^2θsin^2 θ=cos^2 θ + sin^2θcos^2θsin^2 θ+ cos^θsin^2θ =sin^2θcos^2 θ=( ...

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Byju's Answer

Standard XII

Physics

Basic Differentiation Rule

Prove that ...

Question

Prove that $\frac{1 + {tan}^{2} θ}{1 - {cot}^{2} θ} = {(\frac{1 - tan θ}{1 - cot θ})}^{2}$

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Solution

$\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ} = {(\frac{1 - tan θ}{1 - cot θ})}^{2}$

L.H.S
$\frac{1 + {tan}^{2} θ}{1 + {cot}^{2} θ} = \frac{1 + \frac{{sin}^{2} θ}{{cos}^{2} θ}}{1 + \frac{c o s^{2} θ}{s i n^{2} θ}} = \frac{\frac{{cos}^{2} θ + {sin}^{2} θ}{c o s^{2} θ}}{\frac{{sin}^{2} θ + {cos}^{θ}}{{sin}^{2} θ}}$
$= \frac{{sin}^{2} θ}{{cos}^{2} θ} = (tan θ)^{2}$

R.H.S

$= {(\frac{1 - tan θ}{1 - cot θ})}^{2} = {⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{1 - sin θ}{cos θ}}{\frac{1 - cos θ}{sin θ}} ⎞ ⎟ ⎟ ⎟ ⎠}^{2} = {⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{cos θ - sin θ}{cos θ}}{\frac{sin θ - cos θ}{sin θ}} ⎞ ⎟ ⎟ ⎟ ⎠}^{2} = {(\frac{- sin θ}{cos θ})}^{2} = (tan θ)^{2}$

L.H.S=R.H.S

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Similar questions

Prove the following trigonometric identities:

$\frac{1 + t a n^{2} θ}{1 + c o t^{2} θ} = {(\frac{1 - t a n θ}{1 - c o t θ})}^{2} = t a n^{2} θ$

Q. Prove the identity:
(i)

\frac{c o t A + t a n B}{c o t B + t a n A} = c o t A t a n B

(ii)

\frac{1 + t a n^{2} θ}{1 + c o t^{2} θ} = {(\frac{1 - t a n θ}{1 - c o t θ})}^{2}

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Q. Prove that

\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}

, using the identity

{sec}^{2} θ = 1 + {tan}^{2} θ

\frac{1 + \tan^{2} θ}{1 + \cot^{2} θ} = {(\frac{1 + tanθ}{1 - cotθ})}^{2} = \tan^{2} θ

Q. Prove that:

{sec}^{2} θ = 1 + {tan}^{2} θ

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Prove that 1+tan2θ1−cot2θ=(1−tanθ1−cotθ)2

Prove that $\frac{1 + {tan}^{2} θ}{1 - {cot}^{2} θ} = {(\frac{1 - tan θ}{1 - cot θ})}^{2}$