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    JEE Main April Session 2026: Answer Key, Result, Cutoff, Admit Card, Exam Dates, Question Papers

    Trigonometric Ratios of Allied Angles - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • 22 Questions around this concept.

    Solve by difficulty

    $\tan \left(180^{\circ}+\theta\right)+\cot \left(90^{\circ}+\theta\right)=$

    The value of $\tan 1 ^{\circ} \tan 2 ^{\circ} \tan 3 ^{\circ} \ldots \tan 89 ^{\circ} \: \: \: is\\\\$

     

    The value of $\left(\sin 70^{\circ}\right)\left(\cot 10^{\circ} \cot 70^{\circ}-1\right)$ is

    Let $f$ be an odd function defined on the set of real numbers such that for $x \geqslant 0$, $f(x)=3 \sin x+4 \cos x$. Then $f(x)$ at $x=-\frac{11 \pi}{6}$ is equal to:

    The value of $\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}$ is

    Let
    $
    |\cos \theta \cos (60-\theta) \cos (60-\theta)| \leq \frac{1}{8}, \theta \in[0,2 \pi]
    $

    Then, the sum of all $\theta \in[0,2 \pi]$, where $\cos 3 \theta$ attains its maximum value, is :

    Concepts Covered - 2

    Allied Angles (Part 1)

    Allied Angles (Part 1)

    Two angles are called allied if their sum or difference is a multiple of π/2   

    • sin (900 - θ) = cos (θ)

    • cos (900 - θ) = sin (θ)

    • tan (900 - θ) = cot (θ)

    • csc (900 - θ) = sec (θ)          

    • sec (900 - θ) = csc (θ)

    • cot (900 - θ) = tan (θ)

    • sin (900 + θ) = cos (θ)

    • cos (900 + θ) = - sin (θ)

    • tan (900 + θ) = - cot (θ)

    • csc (900 + θ) = sec (θ)          

    • sec (900 + θ) = - csc (θ)

    • cot (900 + θ) = - tan (θ)

    Allied Angles (Part 2)

    Allied Angles (Part 2)

    $\begin{aligned} & \sin \left(180^{\circ}-\theta\right)=\sin (\theta) \\ & \cos \left(180^{\circ}-\theta\right)=-\cos (\theta) \\ & \tan \left(180^{\circ}-\theta\right)=-\tan (\theta) \\ & \csc \left(180^{\circ}-\theta\right)=\csc (\theta) \\ & \sec \left(180^{\circ}-\theta\right)=-\sec (\theta) \\ & \cot \left(180^{\circ}-\theta\right)=-\cot (\theta) \\ & \sin \left(180^{\circ}+\theta\right)=-\sin (\theta) \\ & \cos \left(180^{\circ}+\theta\right)=-\cos (\theta) \\ & \tan \left(180^{\circ}+\theta\right)=\tan (\theta) \\ & \csc \left(180^{\circ}+\theta\right)=-\csc (\theta) \\ & \sec \left(180^{\circ}+\theta\right)=-\sec (\theta) \\ & \cot \left(180^{\circ}+\theta\right)=\cot (\theta)\end{aligned}$

    AID TO REMEMBER

    All the trigonometric functions of a real number of form $2 n(\pi / 2) \pm x(n \in I)$ (i.e. an even multiple of $\pi / 2 \pm x$ ) are numerically equal to the same function of $x$, with sign depending on the quadrant in which terminal side of the angles lies.

    For example $\cos (\pi+x)=\cos (2(\pi / 2)+x)=-\cos (x)$, -ve sign was chosen because $(\pi+x)$ lies in 3rd quadrant and 'cos' is -ve in the third quadrant.

    All the trigonometric functions of a real number of the form $(2 n+1) \pi / 2 \pm x(n \in I)$ (i.e. an even multiple of $\pi / 2 \pm x)$ is numerically equal to the co-function of $x$, with sign depending on the quadrant in which terminal side of the angles lies.

    Note that 'sin' and 'cos' are co-functions of each other, 'tan' and 'cot' are co-functions of each other and 'sec' and 'cosec' are cofunctions of each other.

    For example: $\sec (\pi / 2+x)=-\operatorname{cosec}(x)$, as $(\pi / 2+x)$ lies in the 2 nd quadrant and 'sec' is -ve in $2 n d$ quadrant.

     

    Study it with Videos

    Allied Angles (Part 1)
    Allied Angles (Part 2)

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    Books

    Reference Books

    Allied Angles (Part 1)

    Mathematics for Joint Entrance Examination JEE (Advanced) : Trigonometry

    Page No. : 2.20

    Line : 47

    Allied Angles (Part 2)

    Mathematics for Joint Entrance Examination JEE (Advanced) : Trigonometry

    Page No. : 2.20

    Line : 5

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