JEE MainMathematics

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Trigonometry: JEE Main Mathematics Question with Solution

Q: Why do we discard the root ?

Because for , is strictly between and . The root is negative and outside the range of the sine function.

If the solution of the equation

\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1, x \in\left(0, \frac{\pi}{2}\right)

, is

\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)

, where

\alpha

and

\beta

are integers, then

\alpha+\beta

is equal to :

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5

6

4

3

Step-by-step solutionView

Correct answer

By expressing all logarithms in terms of $\log_{\cos x}(\sin x)$ , the equation yields $\sin x = \frac{-1+\sqrt{5}}{2}$ , giving $\alpha=-1$ , $\beta=5$ , and thus $\alpha+\beta=4$ .

Option analysis

Why each option works or fails

A ·

5

The student identifies $\beta = 5$ from the solution $\sin x = \frac{-1+\sqrt{5}}{2}$ and directly reports $\beta$ instead of computing $\alpha + \beta$ . Check the final quantity requested: the problem asks for $\alpha + \beta$ , where $\alpha = -1$ and $\beta = 5$ , giving $(-1) + 5 = 4$ .

B ·

6

The student incorrectly writes the quadratic equation or misidentifies the root as $\frac{1+\sqrt{5}}{2}$ , leading to $\alpha = 1$ and $\alpha+\beta = 1+5=6$ . Ensure the sign of the linear coefficient in the quadratic equation is handled correctly; for $t^2+t-1=0$ , $\sin x = \frac{-1\pm\sqrt{5}}{2}$ , which means $\alpha = -1$ .

C ·

4

None. The student correctly transforms the equation into a quadratic in $t = \log_{\cos x}(\sin x)$ , solves for $\sin x$ , and calculates $\alpha+\beta = -1+5=4$ . Correctly solved.

D ·

3

The student makes an arithmetic error during the quadratic formula evaluation, computing the discriminant as $b^2-4ac = 1 - (-3) = 4$ or similar, leading to $\beta=4$ and $\alpha+\beta = 3$ . Carefully compute the discriminant: for $t^2+t-1=0$ , $\Delta = 1^2 - 4(1)(-1) = 5$ .

Reviewed route

Solution

StepWorking

01given

Equation $\log_{\cos x} \cot x + 4 \log_{\sin x} \tan x = 1$ with $x \in \left(0, \frac{\pi}{2}\right)$ , and solution $x = \sin^{-1}\left(\frac{\alpha + \sqrt{\beta}}{2}\right)$ where $\alpha, \beta \in \mathbb{Z}$ .

02goal

Find the value of $\alpha + \beta$ .

03approach

Express $\cot x$ as $\frac{\cos x}{\sin x}$ and $\tan x$ as $\frac{\sin x}{\cos x}$ . Use logarithmic quotient rules to convert the equation into a single variable $t = \log_{\cos x} \sin x$ , solve for $t$ , relate $\sin x$ to $\cos x$ , and solve the resulting quadratic in $\sin x$ .

04execute

Expand the terms: $\log_{\cos x} \left(\frac{\cos x}{\sin x}\right) + 4 \log_{\sin x} \left(\frac{\sin x}{\cos x}\right) = 1$ $\implies (1 - \log_{\cos x} \sin x) + 4(1 - \log_{\sin x} \cos x) = 1$ $\implies 5 - \log_{\cos x} \sin x - \frac{4}{\log_{\cos x} \sin x} = 1$ $\implies \log_{\cos x} \sin x + \frac{4}{\log_{\cos x} \sin x} = 4$ .

05execute

Let $t = \log_{\cos x} \sin x$ . Then: $t + \frac{4}{t} = 4 \implies t^2 - 4t + 4 = 0 \implies (t - 2)^2 = 0 \implies t = 2$ .

06execute

Since $t = 2$ , we have $\log_{\cos x} \sin x = 2 \implies \sin x = \cos^2 x = 1 - \sin^2 x$ . Rearranging gives $\sin^2 x + \sin x - 1 = 0$ . By the quadratic formula: $\sin x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$ . Since $x \in \left(0, \frac{\pi}{2}\right)$ , $\sin x > 0$ , so $\sin x = \frac{-1 + \sqrt{5}}{2}$ . Comparing with $\sin^{-1}\left(\frac{\alpha + \sqrt{\beta}}{2}\right)$ , we get $\alpha = -1$ and $\beta = 5$ .

07execute

Compute $\alpha + \beta = -1 + 5 = 4$ .

✓verify

For $\sin x = \frac{\sqrt{5}-1}{2} \approx 0.618$ , $\cos^2 x = 1 - 0.618^2 = 1 - 0.382 = 0.618 = \sin x$ , which confirms $t = 2$ . Both base $\cos x$ and argument $\sin x$ are in $(0, 1)$ , so $\log_{\cos x} \sin x > 0$ , perfectly consistent with $t = 2$ .

Hints that build this answer step by step

If we let $t = \log_{\cos x}(\sin x)$ , how can the equation $\log_{\cos x}\cot x + 4\log_{\sin x}\tan x = 1$ be written in terms of $t$ ?
$\log_{\cos x}(\frac{\cos x}{\sin x}) + 4\log_{\sin x}(\frac{\sin x}{\cos x}) = (1 - t) + 4\left(1 - \frac{1}{t}\right) = 1$
Simplifying $(1 - t) + 4 - \frac{4}{t} = 1$ leads to which values for $t$ ?
$t = 2$ or $t = 2$ (a repeated root $t=2$ )
Using $t = \log_{\cos x}(\sin x) = 2$ , what is the value of $\sin x$ in the interval $\left(0, \frac{\pi}{2}\right)$ ?
$\sin x = \frac{-1 + \sqrt{5}}{2}$
Comparing $\sin x = \frac{-1 + \sqrt{5}}{2}$ to the form $\frac{\alpha + \sqrt{\beta}}{2}$ , what is $\alpha + \beta$ ?
$\alpha = -1, \beta = 5 \implies \alpha + \beta = 4$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we discard the root

\frac{-1-\sqrt{5}}{2}

Because for $x \in (0, \pi/2)$ , $\sin x$ is strictly between $0$ and $1$ . The root $\frac{-1-\sqrt{5}}{2} \approx -1.618$ is negative and outside the range of the sine function.

Answer

By expressing all logarithms in terms of $\log_{\cos x}(\sin x)$ , the equation yields $\sin x = \frac{-1+\sqrt{5}}{2}$ , giving $\alpha=-1$ , $\beta=5$ , and thus $\alpha+\beta=4$ .

Why each option works or fails

A: $5$ - The student identifies $\beta = 5$ from the solution $\sin x = \frac{-1+\sqrt{5}}{2}$ and directly reports $\beta$ instead of computing $\alpha + \beta$ . Check the final quantity requested: the problem asks for $\alpha + \beta$ , where $\alpha = -1$ and $\beta = 5$ , giving $(-1) + 5 = 4$ .
B: $6$ - The student incorrectly writes the quadratic equation or misidentifies the root as $\frac{1+\sqrt{5}}{2}$ , leading to $\alpha = 1$ and $\alpha+\beta = 1+5=6$ . Ensure the sign of the linear coefficient in the quadratic equation is handled correctly; for $t^2+t-1=0$ , $\sin x = \frac{-1\pm\sqrt{5}}{2}$ , which means $\alpha = -1$ .
C · correct: $4$ - None. The student correctly transforms the equation into a quadratic in $t = \log_{\cos x}(\sin x)$ , solves for $\sin x$ , and calculates $\alpha+\beta = -1+5=4$ . Correctly solved.
D: $3$ - The student makes an arithmetic error during the quadratic formula evaluation, computing the discriminant as $b^2-4ac = 1 - (-3) = 4$ or similar, leading to $\beta=4$ and $\alpha+\beta = 3$ . Carefully compute the discriminant: for $t^2+t-1=0$ , $\Delta = 1^2 - 4(1)(-1) = 5$ .

Step-by-step solution

given: Equation $\log_{\cos x} \cot x + 4 \log_{\sin x} \tan x = 1$ with $x \in \left(0, \frac{\pi}{2}\right)$ , and solution $x = \sin^{-1}\left(\frac{\alpha + \sqrt{\beta}}{2}\right)$ where $\alpha, \beta \in \mathbb{Z}$ .
goal: Find the value of $\alpha + \beta$ .
approach: Express $\cot x$ as $\frac{\cos x}{\sin x}$ and $\tan x$ as $\frac{\sin x}{\cos x}$ . Use logarithmic quotient rules to convert the equation into a single variable $t = \log_{\cos x} \sin x$ , solve for $t$ , relate $\sin x$ to $\frac{1+\sqrt{5}}{2}$ , and solve the resulting quadratic in $\sin x$ .
execute: Expand the terms: $\log_{\cos x} \left(\frac{\cos x}{\sin x}\right) + 4 \log_{\sin x} \left(\frac{\sin x}{\cos x}\right) = 1$ $\implies (1 - \log_{\cos x} \sin x) + 4(1 - \log_{\sin x} \cos x) = 1$ $\implies 5 - \log_{\cos x} \sin x - \frac{4}{\log_{\cos x} \sin x} = 1$ $\implies \log_{\cos x} \sin x + \frac{4}{\log_{\cos x} \sin x} = 4$ .
execute: Let $t = \log_{\cos x} \sin x$ . Then: $t + \frac{4}{t} = 4 \implies t^2 - 4t + 4 = 0 \implies (t - 2)^2 = 0 \implies t = 2$ .
execute: Since $t = 2$ , we have $\log_{\cos x} \sin x = 2 \implies \sin x = \cos^2 x = 1 - \sin^2 x$ . Rearranging gives $\sin^2 x + \sin x - 1 = 0$ . By the quadratic formula: $\sin x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$ . Since $x \in \left(0, \frac{\pi}{2}\right)$ , $\sin x > 0$ , so $\sin x = \frac{-1 + \sqrt{5}}{2}$ . Comparing with $\sin^{-1}\left(\frac{\alpha + \sqrt{\beta}}{2}\right)$ , we get $\alpha = -1$ and $\beta = 5$ .
execute: Compute $\alpha + \beta = -1 + 5 = 4$ .
verify: For $\sin x = \frac{\sqrt{5}-1}{2} \approx 0.618$ , $\cos^2 x = 1 - 0.618^2 = 1 - 0.382 = 0.618 = \sin x$ , which confirms $t = 2$ . Both base $\cos x$ and argument $\sin x$ are in $(0, 1)$ , so $\log_{\cos x} \sin x > 0$ , perfectly consistent with $t = 2$ .

Trigonometry: JEE Main Mathematics Question with Solution

What feels right?

By expressing all logarithms in terms of $\log_{\cos x}(\sin x)$ , the equation yields $\sin x = \frac{-1+\sqrt{5}}{2}$ , giving $\alpha=-1$ , $\beta=5$ , and thus $\alpha+\beta=4$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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Trigonometry: JEE Main Mathematics Question with Solution

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