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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:

This question was previously asked in
SSC Selection Post 2021 Question Paper (09-Feb-2022) (Shift-1)
The correct answer is

1

The correct answer is 1.

Tangent of one angle is the cotangent of its complement. Secant of one angle is the cosecant of its complement. These identities allow us to rewrite trigonometric functions of angles greater than 45^\circ in terms of angles less than 45^\circ , which can sometimes make calculations or comparisons easier. In problems like this, recognizing complementary angles helps simplify complex expressions by converting terms into functions of the same angle.

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

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

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

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

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

  8. 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:

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