Trigonometry Previous Year Questions with Solutions

1.0Introduction

Trigonometry is a vital topic in mathematics that deals with the relationships between the angles and sides of triangles. Previous year questions generally focus on concepts like trigonometric ratios, identities, solving equations, and inverse trigonometric functions. These questions help in applying standard formulas such as \sin, \cos, \tan, and their reciprocals, as well as using identities like Pythagorean, angle sum, and difference identities.

Examples often include solving for unknown angles, proving identities, and simplifying trigonometric expressions. Practicing these questions enhances problem-solving skills and provides a deeper understanding of core trigonometric concepts.

2.0Trigonometry Previous Year Questions for JEE with Solutions

JEE Questions in Trigonometry often test advanced concepts related to trigonometric identities, equations, and applications. Some common types of problems include:

• Trigonometric Equations: Questions that require solving equations like \sin x = a, \cos x = b, or more complex trigonometric expressions, including multiple angle and periodicity-based solutions.
• Trigonometric Identities: Problems that involve simplifying or proving standard identities (e.g., sum-to-product, double angle, and reduction formulas) or deriving results using complex identities.
• Solving Triangles: Applications of the Law of Sines and Law of Cosines, along with problems involving the use of trigonometric ratios to solve non-right angled triangles and applying properties of triangles in coordinate geometry.
• Height and Distance Problems: These problems often involve real-world scenarios such as calculating the height of a building, distance between two points, or angles of elevation and depression, frequently requiring the use of advanced trigonometric concepts.
• Multiple Angle and Inverse Trigonometric Functions: Problems that involve inverse functions, transformations of angles (e.g., solving \tan(2x) = 1), or using inverse trigonometric functions to find exact values or angles in specified ranges.

These questions aim to test not only basic knowledge but also the ability to apply advanced techniques and solve complex problems efficiently.

Note: In the JEE Main Mathematics exam, you can generally expect 3 to 5 questions from the Trigonometry chapter, typically ranging from moderate to high difficulty.

3.0Key Concepts for JEE Trigonometry

1. Trigonometric Ratios and Standard Angles:

Key Ratios to Memorize:

$$\begin{gathered} \sin 0^{\circ }=0,\cos 0^{\circ }=1,\tan 0^{\circ }=0 \\ \sin 30^{\circ }=\frac{1}{2},\cos 30^{\circ }=\frac{\sqrt{3}}{2},\tan 30^{\circ }=\frac{1}{\sqrt{3}} \\ \sin 45^{\circ }=\cos 45^{\circ }=\frac{\sqrt{2}}{2},\tan 45^{\circ }=1 \\ \sin 60^{\circ }=\frac{\sqrt{3}}{2},\cos 60^{\circ }=\frac{1}{2},\tan 60^{\circ }=\sqrt{3} \\ \sin 90^{\circ }=1,\cos 90^{\circ }=0,\tan 90^{\circ }=\infty \end{gathered}$$

2. Trigonometric Identities:

Fundamental Identities:

$$\begin{gathered} {\sin }^2\theta +{\cos }^2\theta =1 \\ 1+{\tan }^2\theta ={\sec }^2\theta \\ 1+{\cot }^2\theta ={\csc }^2\theta \end{gathered}$$

Angle Sum and Difference Identities:

$$\begin{gathered} \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A+B)=\cos A\cos B-\sin A\sin B \\ \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \end{gathered}$$

Double Angle and Half Angle Formulas:

$$\begin{gathered} \sin 2A=2\sin A\cos A \\ \cos 2A={\cos }^{2}A-{\sin }^{2}A\text{or}\cos 2A=2{\cos }^{2}A-1 \\ \tan 2A=\frac{2\tan A}{1-{\tan }^{2}A} \\ \sin \frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}} \end{gathered}$$

3. Trigonometric Equations:

General Solutions:

$$\begin{gathered} For\sin \theta =a,\theta ={\sin }^{-1}(a)+2n\pi \text{or}\theta =\pi -{\sin }^{-1}(a)+2n\pi \\ For\cos \theta =a,\theta ={\cos }^{-1}(a)+2n\pi \text{or}\theta =-{\cos }^{-1}(a)+2n\pi \\ For\tan \theta =a,\theta ={\tan }^{-1}(a)+n\pi \end{gathered}$$

• Solving Multiple Angle Equations:

Use period properties of trig functions to find all solutions in the specified range.

4. Inverse Trigonometric Functions:

Key Properties:

$si{n}^{-1}x\text{has range}\left[-\frac{\pi }{2},\frac{\pi }{2}\right]{\cos }^{-1}(x)\text{has range}[0,\pi ],{\tan }^{-1}(x)\text{has range}\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.

Use inverse identities for simplification, e.g., ${\sin }^{-1}(x)+{\cos }^{-1}(x)=\frac{\pi }{2}$.

5. Trigonometric Transformations:

Product-to-Sum and Sum-to-Product Formulas:

$$\begin{gathered} \sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)] \\ \cos A\cos B=\frac{1}{2}[\cos (A+B)+\cos (A-B)] \\ \sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)] \end{gathered}$$

Transforming expressions for simplification in integrals or solving equations.

6. Trigonometric Graphs:

• Understand amplitude, period, and phase shift for sine, cosine, and tangent functions.
• Recognize transformations of graphs of functions like , where A affects amplitude, B affects period, and C affects phase shift.
• Be comfortable analyzing and sketching graphs of composite trigonometric functions.

7. Heights and Distances Problems:

• Use trigonometric ratios to solve real-life problems, such as finding the height of a building or the distance between two points based on angles of elevation and depression.
• Key Formula:

Height =distance x tan(angle of elevation)

8. Applications to Solving Triangles:

Law of Sines and Law of Cosines for solving non-right angled triangles.

$$\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\text{(Law of Sines)} \\ {c}^2=a^2+b^2-2ab\cos C\text{(Law of Cosines)} \end{gathered}$$

9. Solving Trigonometric Inequalities:

• Solve inequalities like $\sin x>\frac{1}{2}or\cos x\le 0$ by considering the range of values of trigonometric functions within given intervals.

4.0JEE Mains Past Year Questions with Solutions on Trigonometry

Trignometry

1. The number of solutions of sin²x + (2 + 2x – x²)sinx – 3(x – 1)² = 0, where –π ≤ x ≤ π, is

Ans. (2)

Sol. sin²x – (x² – 2x – 2)sinx – 3(x – 1) ² = 0 

sin²x – (x – 1)²)sinx – 3(x–1)² = 0

roots :

sinx = –3 (reject) or (x – 1)²

sinx = (x – 1)²

2. Let $\left|\cos \theta \cos (60-\theta )\cos (60-\theta )\right|\le \frac{1}{8},\theta \in [0,2\pi ]$_. Then, the sum of all $\theta \in [0,2\pi ]$ , where cos 3θ attains its maximum value, is :

(1) 9π

(2) 18 π

3) 6 π

(4) 15 π

Ans. (3)

Sol. We know that 

$$\begin{gathered} (\cos \theta )(\cos (60^{\circ }-\theta ))(\cos (60^{\circ }+\theta ))=\frac{1}{4}\cos 3\theta \\ \text{So equation reduces to}\left|\frac{1}{4}\cos 3\theta \right|\le \frac{1}{8} \\ \Rightarrow |\cos 3\theta |\le \frac{1}{2} \\ \Rightarrow -\frac{1}{2}\le \cos 3\theta \le \frac{1}{2} \\ \Rightarrow -\frac{1}{2}\le \cos 3\theta \le \frac{1}{2} \\ \text{Maximum value}of\cos 3\theta =\frac{1}{2} \\ \Rightarrow 3\theta =2n\pi \pm \frac{\pi }{3} \\ \Rightarrow \theta =\frac{2n\pi \pm \frac{\pi }{3}}{3}=\frac{2n\pi }{3}\pm \frac{\pi }{9} \\ \text{As }\theta \in [0,2\pi ],thevalidvaluesof\theta are: \\ \theta =\left\{\frac{\pi }{9},\frac{5\pi }{9},\frac{7\pi }{9},\frac{11\pi }{9},\frac{13\pi }{9},\frac{17\pi }{9}\right\} \\ \text{Sum of values:} \\ \frac{\pi }{9}+\frac{5\pi }{9}+\frac{7\pi }{9}+\frac{11\pi }{9}+\frac{13\pi }{9}+\frac{17\pi }{9} \\ =\frac{54\pi }{9}=6\pi \\ \text{Final Answer:}\boxed{6\pi } \end{gathered}$$

3. If $2{\tan }^2\theta -5\sec \theta =1$ has exactly 7 solutions in the interval $\left[0,\frac{n\pi }{2}\right]$, for the least value of $n\in \mathbb{N}then{\sum }_{k=1}^n\frac{k}{2^k}$ is equal to : 

$$\begin{gathered} (1)\frac{1}{2^{15}}(2^{14}-14) \\ (2)\frac{1}{2^{14}}(2^{15}-15) \\ (3)1-\frac{15}{2^{13}} \\ (4)\frac{1}{2^{13}}(2^{14}-15) \end{gathered}$$

Ans. (4)

Sol.

$$\begin{gathered} 2{\tan }^2\theta -5\sec \theta -1=0 \\ \Rightarrow 2{\sec }^2\theta -5\sec \theta -3=0 \\ \Rightarrow (2\sec \theta +1)(\sec \theta -3)=0 \\ \Rightarrow \sec \theta =-\frac{1}{2}\text{(not valid)}\text{or}\sec \theta =3 \\ \Rightarrow \cos \theta =\frac{1}{3} \\ Now,\cos \theta =\frac{1}{3}hasexactly7solutionsin\left[0,\frac{n\pi }{2}\right]whenn=13. \\ \text{So,} \\ {\sum }_{k=1}^{13}\frac{k}{2^k}=S \\ S=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots +\frac{12}{2^{13}}+\frac{13}{2^{14}} \\ \text{Let us find using a known identity for this summation:} \\ S={\sum }_{k=1}^n\frac{k}{2^k} \\ \Rightarrow S=\frac{a-(n+1)}{(1-r{)}^2}+\frac{(n+1)}{2^n} \\ \text{Alternatively, as derived in the image:} \\ \frac{S}{2}=\frac{1}{2^2}+\frac{2}{2^3}+\cdots +\frac{13}{2^{14}} \\ \Rightarrow S-\frac{S}{2}=\frac{1}{2}+\frac{1}{2^2}+\cdots +\frac{12}{2^{13}}+\frac{13}{2^{14}} \\ \Rightarrow \frac{S}{2}=1-\frac{1}{2^{13}}-\frac{13}{2^{14}} \\ \Rightarrow S=2\left(1-\frac{1}{2^{13}}-\frac{13}{2^{14}}\right) \\ \text{Now express with common denominator:} \\ S=2\left(\frac{2^{14}-2-13}{2^{14}}\right)=\frac{2(2^{14}-15)}{2^{14}}=\frac{2^{15}-30}{2^{14}}=\frac{2^{14}-15}{2^{13}} \end{gathered}$$

4. If 2sin³x + sin 2x cos x + 4sinx – 4 = 0 has exactly 3 solutions in the interval $\left[0,\frac{n\pi }{2}\right],n\in \mathbb{N}$ , then the roots of the equation $x^2+nx+(n-3)=0$ belong to :

$$\begin{gathered} (1)(0,\infty ) \\ (2)(-\infty ,0) \\ (3)\left(\frac{-\sqrt{17}}{2},\frac{\sqrt{17}}{2}\right) \\ (4)Z \end{gathered}$$

Ans. (2)

Sol.

$$\begin{gathered} 2{\sin }^{3}x+2\sin x{\cos }^{2}x+4\sin x-4=0 \\ 2{\sin }^{3}x+2\sin x(1-{\sin }^{2}x)+4\sin x-4=0 \\ 6\sin x-4=0 \\ \sin x=\frac{2}{3} \\ \text{n = 5 (in the given interval)} \\ {x}^2+5x+2=0 \\ x=\frac{-5\pm \sqrt{17}}{2} \\ \text{Required interval:}(-\infty ,0) \end{gathered}$$

5.0Inverse Trigonometric Function

1. Given the inverse trigonometric function assumes principal values only. Let x, y be any two real numbers in [–1,1] such that cos^–1x – sin^–1 y = a, $-\frac{\pi }{2}\le a\le \pi$. Then, the minimum value of x² + y² + 2xy sina is

(1) –1

(2) 0

(3) $\frac{-1}{2}$

(4) $\frac{1}{2}$

Ans. (2)

Sol.

$$\begin{gathered} {\cos }^{-1}x-\left(\frac{\pi }{2}-{\cos }^{-1}y\right)=a \\ {\cos }^{-1}x+{\cos }^{-1}y=\frac{\pi }{2}+a \\ \alpha \in \left[\frac{\pi }{2},\pi \right],\frac{\pi }{2}+\alpha \in \left[0,\frac{3\pi }{2}\right] \\ {\cos }^{-1}{\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right)}^2=\frac{\pi }{2}+\alpha \\ xy-\sqrt{1-x^2}\sqrt{1-y^2}=-\sin \alpha \\ (xy+\sin \alpha {)}^2=(1-x^2)(1-y^2) \\ {x}^2{y}^2+2xy\sin \alpha +{\sin }^2\alpha =1-x^2-y^2+x^2{y}^2 \\ {x}^2+y^2+2xy\sin \alpha =1-{\sin }^2\alpha \\ {x}^2+y^2+2xy\sin \alpha ={\cos }^2\alpha \\ \text{Minimum value of }{\cos }^2\alpha =0\text{ at }\alpha =\frac{\pi }{2} \\ \text{Option (2) is correct} \end{gathered}$$

2. For n Î N, if cot^–13 + cot^–14 + cot^–15 + ${\cot }^{-1}$ $n=\frac{\pi }{4}$ , then n is equal to ________.    

Ans. (47)

Sol.

$$\begin{gathered} {\cot }^{-1}3+{\cot }^{-1}4+{\cot }^{-1}5+{\cot }^{-1}n=\frac{\pi }{4} \\ {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{n}\right)=\frac{\pi }{4} \\ {\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)\text{ when }ab<1 \\ {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{4}\right)={\tan }^{-1}\left(\frac{\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{3}\cdot \frac{1}{4}}\right)={\tan }^{-1}\left(\frac{7/12}{11/12}\right)={\tan }^{-1}\left(\frac{7}{11}\right) \\ {\tan }^{-1}\left(\frac{7}{11}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)={\tan }^{-1}\left(\frac{\frac{7}{11}+\frac{1}{5}}{1-\frac{7}{11}\cdot \frac{1}{5}}\right)={\tan }^{-1}\left(\frac{82/55}{48/55}\right)={\tan }^{-1}\left(\frac{82}{48}=\frac{41}{24}\right) \\ {\tan }^{-1}\left(\frac{41}{24}\right)+{\tan }^{-1}\left(\frac{1}{n}\right)=\frac{\pi }{4} \\ ta{n}^{-1}\left(\frac{1}{n}\right)=\frac{\pi }{4}-{\tan }^{-1}\left(\frac{41}{24}\right) \\ {\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right) \\ {\tan }^{-1}\left(\frac{1}{n}\right)={\tan }^{-1}\left(\frac{1-\frac{41}{24}}{1+\frac{41}{24}}\right)={\tan }^{-1}\left(\frac{-17/24}{65/24}\right)={\tan }^{-1}\left(\frac{-17}{65}\right) \\ \frac{1}{n}=\frac{1}{47}\Rightarrow n=47 \\ \text{n = 47} \end{gathered}$$

3. If the domain of the function $f(x)={\sin }^{-1}\left(\frac{x-1}{2x+3}\right)$ is R – (α, β) then 12αβ is equal to :

(1) 36

(2) 24

(3) 40

(4) 32

Ans. (4)

Sol.

$$\begin{gathered} \text{Domain of }f(x)={\sin }^{-1}\left(\frac{x-1}{2x+3}\right)is: \\ 2x+3\ne 0x\ne -\frac{3}{2}\text{and}\left|\frac{x-1}{2x+3}\right|\le 1 \\ \left|x-1\right|\le \left|2x+3\right| \end{gathered}$$

$$\begin{gathered} For\left|2x+3\right|\ge \left|x-1\right|: \\ x\in (-\infty ,-4)\cup \left(-\frac{2}{3},\infty \right) \\ So,\alpha =-4,\beta =-\frac{2}{3}\Rightarrow 12\alpha \beta =12(-4)\left(-\frac{2}{3}\right)=32 \end{gathered}$$

4. If $a={\sin }^{-1}(\sin (5))andb={\cos }^{-1}(\cos (5))$, then $a^2+b^2$ is equal to

$$\begin{gathered} (1)4{\pi }^2+25 \\ (2)8{\pi }^2-40\pi +50 \\ (3)4{\pi }^2-20\pi +50 \\ (4)25 \end{gathered}$$

Ans. (2)

Sol.

$$\begin{gathered} a={\sin }^{-1}(\sin (5))=5-2\pi \text{(since range of }{\sin }^{-1}\text{ is }[-\frac{\pi }{2},\frac{\pi }{2}]) \\ b={\cos }^{-1}(\cos (5))=5\text{(since range of }{\cos }^{-1}\text{ is }[0,\pi ]) \\ {a}^2+b^2=(5-2\pi {)}^2+5^2 \\ =25-20\pi +4{\pi }^2+25 \\ =50-20\pi +4{\pi }^2 \\ =8{\pi }^2-40\pi +50 \end{gathered}$$

Trigonometric Ratio and identity

1. Let S = {sin²2θ : (sin⁴ θ + cos⁴ θ)x² + (sin2 θ)x + (sin⁶ θ + cos⁶ θ) = 0 has real roots}. If α and β be the smallest and largest elements of the set S, respectively, then 3((α – 2)² + (β – 1)²) equals….. 

Ans. (4)

Sol.

$$\begin{gathered} D=(\sin 2\theta {)}^2-4\left(1-\frac{{\sin }^{2}2\theta }{2}\right)\left(1-\frac{3}{4}{\sin }^{2}2\theta \right) \\ =(\sin 2\theta {)}^2-4\left(1-\frac{5}{4}{\sin }^{2}2\theta +\frac{3}{8}{\sin }^{4}2\theta \right) \\ D=-\frac{3}{2}{\sin }^{4}2\theta +6{\sin }^{2}2\theta -4>0 \\ 3{\sin }^{4}2\theta -12{\sin }^{2}2\theta +8<0 \\ {\sin }^{2}2\theta =\frac{12\pm \sqrt{122-12.8}}{6}=\frac{12\pm 4\sqrt{3}}{6}=\frac{6\pm 2\sqrt{3}}{3} \\ {\sin }^{2}2\theta =2\pm \frac{2}{\sqrt{3}},\text{but }{\sin }^{2}2\theta \in [0,1] \\ \therefore \alpha =2-\frac{2}{\sqrt{3}},\beta =1\Rightarrow (\alpha -2{)}^2=\frac{4}{3},(\beta -1{)}^2=0 \\ 3(\alpha -2{)}^2+(\beta -1{)}^2=4 \end{gathered}$$

2. Suppose $\theta \in \left[0,\frac{\pi }{4}\right]$is a solution of 4 cosθ – 3 sinθ = 1. Then cosθ is equal to : 

$$\begin{gathered} (1)\frac{4}{3(\sqrt{6}-2)} \\ (2)\frac{6-\sqrt{6}}{3(\sqrt{6}-2)} \\ (3)\frac{6+\sqrt{6}}{3(\sqrt{6}+2)} \\ (4)\frac{4}{3(\sqrt{6}+2)} \end{gathered}$$

 Ans. (1)

Sol.

$$\begin{gathered} 4\left(\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)-3\left(\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)=1 \\ \text{Given: }\tan \frac{\theta }{2}=t \\ \Rightarrow \frac{4(1-t^2)-6t}{1+t^2}=1 \\ \Rightarrow 4-4t^2-6t=1+t^2 \\ \Rightarrow 5t^2+6t-3=0 \\ \Rightarrow t=\frac{-6\pm \sqrt{36-4(5)(-3)}}{2(5)} \\ =\frac{-6\pm \sqrt{96}}{10} \\ =\frac{-6\pm 4\sqrt{6}}{10} \\ t=\frac{-3+2\sqrt{6}}{5} \\ \cos \theta =\frac{1-t^2}{1+t^2}=\frac{1-{\left(\frac{2\sqrt{6}-3}{5}\right)}^2}{1+{\left(\frac{2\sqrt{6}-3}{5}\right)}^2}=\frac{1-\left(\frac{24+9-12\sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12\sqrt{6}}{25}\right)} \\ =\frac{25-33+12\sqrt{6}}{25+33-12\sqrt{6}}=\frac{12\sqrt{6}-8}{58-12\sqrt{6}}=\frac{6\sqrt{6}-4}{29-6\sqrt{6}}\times \frac{29+6\sqrt{6}}{29+6\sqrt{6}} \\ =\frac{100+150\sqrt{6}}{625}=\frac{4+6\sqrt{6}}{25}\times \frac{4-6\sqrt{6}}{4-6\sqrt{6}} \\ =\frac{-200}{25(4-6\sqrt{6})}=\frac{-8}{4-6\sqrt{6}}=\frac{4}{3\sqrt{6}-2} \end{gathered}$$

3. Let the range of the function $f(x)=f(x)=\frac{1}{2+\sin 3x+\cos 3x},x\in \mathbb{R}$, x IR be [a, b]. If α and β are respectively the A.M. and the G.M. of a and b, then is equal to :

$$\begin{gathered} (1)\sqrt{2} \\ (2)2 \\ (3)\sqrt{\pi } \\ (4)\square \end{gathered}$$

Ans. (1)

Sol.  

$$\begin{gathered} f(x)\frac{1}{2+\sin 3x+\cos 3x} \\ \left[\frac{1}{2+\sqrt{2}},\frac{1}{2-\sqrt{2}}\right] \\ \frac{\alpha }{\beta }=\frac{a+b}{2} \\ \sqrt{ab}=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right) \\ =\frac{1}{2}\left(\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}+\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}\right) \\ =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{2\times \sqrt{2}}=\sqrt{2} \end{gathered}$$

4. Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a + b)² is equal to : 

(1) 72

(2) 60 

(3) 80

(4) 64 

Ans. (1 )

Sol.

Area = (4cosθ + 2sinθ) (2cosθ + 4sinθ)

= 8cos²θ + 16sinθcosθ + 4sinθcosθ + 8sin²θ

= 8 + 20 sinθcosθ

= 8 + 10 sin2θ

Max Area = 8 + 10 = 18 (sin2θ = 1) θ = 45°

(a + b)² = (4cosθ + 2sinθ + 2cosθ + 4sinθ)²

= (6cosθ + 6sinθ)²

= 36 (sinθ + cosθ)²

= $36({\sqrt{2}}^2)$

= 72

5. If tanA =  $\mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}},\tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}\tan C={\left(x^{-3}+x^{-2}+x^{-1}\right)}^{\frac{1}{2}},0<A,B,C<\frac{\pi }{2},then$  A + B is equal to :

$$\begin{gathered} (1)C \\ (2)\pi -C \\ (3)2\pi -C \\ (4)\frac{\pi }{2}-C \end{gathered}$$

Ans. (1)

Sol.

Finding tan (A + B) we get

$$\begin{gathered} \frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{1}{\sqrt{x\left(x^2+x+1\right)}}+\frac{\sqrt{x}}{\sqrt{x^2+x+1}}}{1-\frac{1}{x^2+x+1}} \\ \Rightarrow \tan (\mathrm{A}+\mathrm{B})=\frac{(1+x)\left(\sqrt{x^2+x+1}\right)}{\left(x^2+x\right)(\sqrt{x})} \\ \frac{(1+x)\left(\sqrt{x^2+x+1}\right)}{\left(x^2+x\right)(\sqrt{x})} \\ \tan (A+B)=\frac{\sqrt{x^2+x+1}}{x\sqrt{x}}=\tan C \\ A+B=C \end{gathered}$$