If tan-t then tan 2 - 2 - is equal to

The correct option is A 1+t/1 t 1+t/1 t tan 2 θ + sec 2 θ =2 tan θ/1 tan^2 θ + 1+tan^2 θ/1 tan^2 θ =2 tan θ + 1 + tan^2 θ/1 tan^2 θ =(1+tan θ)^2/1 tan^2 θ = ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If tanθ=t the...

Question

If $t a n θ = t$ then $t a n 2 θ + s e c 2 θ$ is equal to

$\frac{1 + t}{1 - t}$

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$\frac{1 - t}{1 + t}$

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$\frac{2 t}{1 - t}$

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$\frac{2 t}{1 + t}$

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Solution

The correct option is A
$\frac{1 + t}{1 - t}$

$\frac{1 + t}{1 - t}$

$t a n 2 θ + s e c 2 θ = \frac{2 t a n θ}{1 - t a n^{2} θ} + \frac{1 + t a n^{2} θ}{1 - t a n^{2} θ} = \frac{2 t a n θ + 1 + t a n^{2} θ}{1 - t a n^{2} θ} = \frac{(1 + t a n θ)^{2}}{1 - t a n^{2} θ} = \frac{(1 + t a n θ) (1 + t a n θ)}{(1 + t a n θ) (1 - t a n θ)} = \frac{1 + t a n θ}{1 - t a n θ} = \frac{1 + t}{1 - t} [t a n θ = t (given)]$

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Similar questions

Q. If

tan θ + cot θ = 5

, then

{tan}^{2} θ + {cot}^{2} θ =

If (tan $θ$ + cot $θ$ ) = 5 then ( $t a n^{2} θ$ + $c o t^{2} θ$ ) = ?

(a) 27 (b) 25 (c) 24 (d) 23

Q. If

\tan θ = t

then

\tan 2 θ + \sec 2 θ =

(a)

\frac{1 + t}{1 - t}

(b)

\frac{1 - t}{1 + t}

(c)

\frac{2 t}{1 - t}

(d)

\frac{2 t}{1 + t}

Q. If (tan θ + cot θ) = 5 then (tan² θ + cot² θ) = ?
(a) 27
(b) 25
(c) 24
(d) 23

Q. Prove that

\frac{\tan θ - \cot θ}{\sin θ \cos θ} = \tan^{2} θ - \cot^{2} θ

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If tan θ=t then tan 2θ+sec 2θ is equal to

The correct option is A 1+t1−t 1+t1−t tan 2θ+sec 2θ=2 tan θ1−tan2 θ+1+tan2θ1−tan2θ=2 tan θ+1+tan2 θ1−tan2θ=(1+tan θ)21−tan2 θ=(1+tan θ)(1+tan θ)(1+tan θ)(1−tan θ)=1+tan θ1−tan θ=1+t1−t [tan θ=t(given)]

If $t a n θ = t$ then $t a n 2 θ + s e c 2 θ$ is equal to