The general solutions of the equation ² θ+ 2 θ=1 are

Correct Answer is : nπ,nπ±π3,n∈Z{n} π, {n} π ± π/3, {n} {Z}nπ,nπ±3π,n∈Z

{n} π, {n} π ± π/3, {n} {Z}

{n} π, {n} π ± π/4, {n} {Z}

{{n} π}{4}, {{n} π}{4} ± π/3, {n} {Z}

{n} π, {n} π ± π/6, {n} {Z}

Correct Answer is : nπ,nπ±π3,n∈Z{n} π, {n} π ± π/3, {n} {Z}nπ,nπ±3π,n∈Z

Test Index

MHTCET 5 May 2025 Shift 2 Paper

Question : 146 of 150

Marks: +1, -0

\tan ^{2} \theta+\sec 2 \theta=1

are

$\mathrm{n} \pi, \mathrm{n} \pi \pm \frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}$
$\mathrm{n} \pi, \mathrm{n} \pi \pm \frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
$\quad \frac{\mathrm{n} \pi}{4}, \frac{\mathrm{n} \pi}{4} \pm \frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}$
$\quad \mathrm{n} \pi, \mathrm{n} \pi \pm \frac{\pi}{6}, \mathrm{n} \in \mathbb{Z}$

Solution:

Concept:

Convert

\sec 2\theta

into terms of

\tan\theta

and solve the resulting quadratic in

\tan^2\theta

Explanation:

Use the identity

\sec 2\theta = \dfrac{1+\tan^2\theta}{1-\tan^2\theta}

Substitute in the given equation:

\tan^2\theta + \frac{1+\tan^2\theta}{1-\tan^2\theta}=1

Let

x = \tan^2\theta

. Then:

x + \frac{1+x}{1-x}=1

Multiply both sides by

1-x

x(1-x)+1+x=1-x

Simplify:

3x - x^2 = 0

x(3-x)=0

\tan^2\theta = 0

\tan^2\theta = 3

For

\tan\theta = 0

\theta = n\pi

For

\tan^2\theta = \tan^2\frac{\pi}{3}

\theta = n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}

Answer:

\theta = n\pi, \ n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}

Hence, option A is correct.

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