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If tan 15θ = cot 15θ (0° < θ < 10°) , then the value of θ is:

This question was previously asked in
SSC Selection Post 2023 Graduation Level Question Paper (30 Jun, 2023) (Shift 1)
The correct answer is

3°

The correct answer is 3°.

Given the options, and the common practice in exams, the value derived from the $45^\circ$ case is often the expected answer. Final Answer Determination The value $\theta = 3^\circ$ satisfies the equation $\tan(15\theta) = \cot(15\theta)$ because $\tan(15 \times 3^\circ) = \tan(45^\circ) = 1$ and $\cot(15 \times 3^\circ) = \cot(45^\circ) = 1$. Furthermore, $3^\circ$ is within the specified range $0^\circ < \theta < 10^\circ$. This corresponds to one of the provided options.

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Similar Questions

  1. If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.

  2. The value of cot (-315°) is :

  3. The value of \(\frac{\tan^2 (22^\circ - \theta)-\tan (\theta + 68^\circ)\ -\ \text{cosec}^2 (68^\circ + \theta)+\cot (22^\circ - \theta)} {3 (\cot^2 52^\circ - \sec^2 38^\circ)+ 2 (\text{cosec}^2 \ 28^\circ - \tan^2 62^\circ)}\)  is:

  4. The value of \(\frac{{{{\sec }^2}60^\circ {{\cos }^2}45^\circ \, + \,{\rm{cose}}{{\rm{c}}^2}30^\circ }}{{{{\sec }^2}45^\circ - \tan 45^\circ }}\)  is:

  5. The value of \(\sin^2\frac{2\pi}{3}+\cos^2\frac{5\pi}{6}-\tan^2\frac{3\pi}{4}\)  is :

  6. sinθ + cosec θ = 2, then the value of sin 99 θ + cosec 99 θ is:

  7. If cos θ = \(\frac{5}{13}\)  , what is the value of cot θ ?

  8. Determine the largest angle if the angles of a triangle are in the ratio 3 : 4 : 5.

  9. The value of the expression \(\rm \frac{4\sin^230^{\circ}+\cos^260^{\circ}-\tan^245^{\circ}}{2\sin60^{\circ}\cos30^{\circ}-\tan45^{\circ}}\)  is:

  10. The value of cosec 41° - sec 49° is:


Important Questions from Circular Measure of Angles

  1. If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.

  2. 1 + tan 15° cot 75° is equal to:

  3. The value of cot (-315°) is :

  4. θ = 135°; γ = 15°.

    What is the value of 2 cos(θ) sin(γ)?

  5. If tan(90 – θ) = \(\frac{2}{\sqrt3}\), then the value of \(2\sqrt3\) tan θ + 1 is:

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