Trigonometric Ratios & Identities JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Trigonometric Ratios & Identities, free to read — no sign-in needed. The full chapter has 9 questions; sign in to attempt the remaining 5 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   Let $\alpha=\frac{1}{\sin 60^{\circ} \sin 61^{\circ}}+\frac{1}{\sin 62^{\circ} \sin 63^{\circ}}+\ldots+\frac{1}{\sin 118^{\circ} \sin 119^{\circ}}$ Then the value of $\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2$ is __________
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   Answer: 3

   $\begin{aligned} & \alpha=\sum_{\mathrm{r}-30}^{59} \frac{1}{\sin (2 \mathrm{r})^{\circ} \sin (2 \mathrm{r}+1)^{\circ}} \\ & \frac{\alpha}{\operatorname{cosec} 1^{\circ}}=\sum_{\mathrm{r}-30}^{59} \frac{\sin 1^{\circ}}{\sin (2 \mathrm{r})^{\circ} \sin (2 \mathrm{r}+1)^{\circ}} \\ & =\sum_{\mathrm{r}-30}^{59}\left(\cot (2 \mathrm{r})^{\circ}-\cot (2 \mathrm{r}+1)^{\circ}\right) \\ & =\cot 60^{\circ}-\cot 61^{\circ} \\ & \quad+\cot 62^{\circ}-\cot 63^{\circ} \\ & \quad \vdots \\ & +\cot \left(118^{\circ}\right)-\cot \left(119^{\circ}\right) \\ & \frac{\alpha}{\operatorname{cosec} 1^{\circ}}=\cot 60^{\circ}=\frac{1}{\sqrt{3}} \\ & \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2=3\end{aligned}$
2. Q2JEE Advanced Adv 2022 (Paper 2)
   Let $\alpha$ and $\beta$ be real numbers such that $-\dfrac{\pi }{4}\lt \beta \lt 0\lt \alpha \lt \dfrac{\pi }{4}$. If $\sin \left(\alpha +\beta \right)=\dfrac{1}{3}$ and $\cos \left(\alpha -\beta \right)=\dfrac{2}{3}$, then the greatest integer less than or equal to ${\left(\dfrac{\sin \alpha }{\cos \beta }+\dfrac{\cos \beta }{\sin \alpha }+\dfrac{\cos \alpha }{\sin \beta }+\dfrac{\sin \beta }{\cos \alpha }\right)}^{2}$ is
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   Answer: 1

   ${\left(\dfrac{\sin \alpha }{\cos \beta }+\dfrac{\cos \alpha }{\sin \beta }+\dfrac{\cos \beta }{\sin \alpha }+\dfrac{\sin \beta }{\cos \alpha }\right)}^{2}$ $={\left(\dfrac{\cos \left(\alpha -\beta \right)}{\sin \beta \cos \beta }+\dfrac{\cos \left(\alpha -\beta \right)}{\sin \alpha \cos \alpha }\right)}^{2}$ $={\left(2\cos \left(\alpha -\beta \right)\left\{\dfrac{1}{\sin 2\beta }+\dfrac{1}{\sin 2\alpha }\right\}\right)}^{2}$ $=\dfrac{16}{9}{\left(\dfrac{2\sin \left(\alpha +\beta \right)\cdot \cos \left(\alpha -\beta \right)}{\sin 2\alpha \cdot \sin 2\beta }\right)}^{2}$ $=\dfrac{16}{9}{\left(\dfrac{4\cdot \dfrac{1}{3}\cdot \dfrac{2}{3}}{\cos \left(2\alpha -2\beta \right)-\cos \left(2\alpha +2\beta \right)}\right)}^{2}$ $=\dfrac{16}{9}{\left(\dfrac{\dfrac{8}{9}}{\left(2{\cos }^{2}\left(\alpha -\beta \right)-1\right)-\left(1-2{\sin }^{2}\left(\alpha +\beta \right)\right)}\right)}^{2}$ $=\dfrac{16}{9}{\left(\dfrac{\dfrac{8}{9}}{\dfrac{8}{9}-2+\dfrac{2}{9}}\right)}^{2}$ $=\dfrac{16}{9}$ i.e. $\left[\dfrac{16}{9}\right]=1$
3. Q3JEE Advanced Adv 2019 (Paper 2)
   The value of ${\sec }^{-1}⁡\left(\dfrac{1}{4}\sum _{k=0}^{10}\sec ⁡\left(\dfrac{7\pi }{12}+\dfrac{k\pi }{2}\right)\sec ⁡\left(\dfrac{7\pi }{12}+\dfrac{\left(k+1\right)\pi }{2}\right)\right)$ in the interval $\left[-\dfrac{\pi }{4},\dfrac{3\pi }{4}\right]$ equals
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   Answer: 0

   ${\text{sec}}^{-1}\left[\dfrac{1}{4}\sum _{k=0}^{10}\sec \left(\dfrac{7\pi }{12}+\dfrac{k\pi }{2}\right)\sec \left(\dfrac{7\pi }{12}+\dfrac{k\pi }{2}+\dfrac{\pi }{2}\right)\right]$ $\begin{matrix}={\text{sec}}^{-1}\left[\dfrac{-1}{4}\sum _{k=0}^{10}\sec \left(\dfrac{7\pi }{12}+\dfrac{k\pi }{2}\right)\text{cosec}\left(\dfrac{7\pi }{12}+\dfrac{k\pi }{2}\right)\right] & \left(∴\sec \left(\dfrac{\pi }{2}+\theta \right)=-\text{cosec}\theta \right)\end{matrix}$ $={sec}^{-1}\left(-\dfrac{1}{4}\sum _{k=0}^{10}\dfrac{1}{\cos ⁡\left(\dfrac{7\pi }{12}+\dfrac{k\pi }{2}\right)\sin ⁡\left(\dfrac{7\pi }{12}+\dfrac{k\pi }{2}\right)}\right)$ $\begin{matrix}={\sec }^{-1}⁡\left(-\dfrac{1}{4}\sum _{k=0}^{10}\dfrac{2}{\sin ⁡\left(\dfrac{7\pi }{6}+k\pi \right)}\right) & \left(∴2\sin ⁡\theta \cos ⁡\theta =\sin ⁡2\theta \right)\end{matrix}$ now if $\begin{matrix}k=even & \sin ⁡\left(2n\pi +\dfrac{7\pi }{6}\right)=\sin ⁡\dfrac{7\pi }{6}=-\dfrac{1}{2}\end{matrix}$ and if $\begin{matrix}k=odd & \sin ⁡\left(\left(2n+1\right)\pi +\dfrac{7\pi }{6}\right)=-\sin ⁡\dfrac{7\pi }{6}=\dfrac{1}{2}\end{matrix}$ hence $={\sec }^{-1}⁡\left(-\dfrac{1}{2}\left(\dfrac{1}{-\dfrac{1}{2}}+\dfrac{1}{\dfrac{1}{2}}+\dfrac{1}{-\dfrac{1}{2}}+\ldots ..\right)\right)$ $={\sec }^{-1}⁡1$ $=0$
4. Q4JEE Advanced Adv 2018 (Paper 1)
   Let $a,b,c$ be three non-zero real numbers such that the equation $\sqrt{3}a\cos ⁡x+2b\sin ⁡x=c,x\in \left[-\dfrac{\pi }{2},\dfrac{\pi }{2}\right]$ has two distinct real roots $\alpha$ and $\beta$ with $\alpha +\beta =\dfrac{\pi }{3}$. Then, the value of $\dfrac{b}{a}$ is
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   Answer: 0.5

   $\sqrt{3}\cos ⁡x+\dfrac{2b}{a}\sin ⁡x=\dfrac{c}{a}$ Now, $\sqrt{3}\cos ⁡\alpha +\dfrac{2b}{a}\sin ⁡\alpha =\dfrac{c}{a}\ldots \left(i\right)$ $\sqrt{3}\cos ⁡\beta +\dfrac{2b}{a}\sin ⁡\beta =\dfrac{c}{a}\ldots \left(ii\right)$ From equations $\left(i\right)$ and $\left(ii\right)$ $\Rightarrow \sqrt{3}\left[\cos ⁡\alpha -\cos ⁡\beta \right]+\dfrac{2b}{a}\left(\sin ⁡\alpha -\sin ⁡\beta \right)=0$ $\Rightarrow \sqrt{3}\left[-2\sin ⁡\left(\dfrac{\alpha +\beta }{2}\right)\sin ⁡\left(\dfrac{\alpha -\beta }{2}\right)\right]+\dfrac{2b}{a}\left[2\cos ⁡\left(\dfrac{\alpha +\beta }{2}\right)\sin ⁡\left(\dfrac{\alpha -\beta }{2}\right)\right]=0$ $\Rightarrow -\sqrt{3}+2\sqrt{3}\cdot \dfrac{b}{a}=0$ $\Rightarrow \dfrac{b}{a}=\dfrac{1}{2}=0.5$