Trigonometric Functions Test 23

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Trigonometric Functions Test 23

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Weekly Quiz Competition

Question 1
1 / -0

The general solution of the trigonometric equation $$\tan^2 \theta = 1$$ is ______

A
$$\theta = n \pi \pm \dfrac{\pi}{4}, n \in z$$

B
$$\theta = n \pi, n \in z$$

C
$$\theta = n \pi \pm \dfrac{\pi}{3}, n \in z$$

D
$$\theta = n \pi \pm \dfrac{\pi}{6}, n$$

Solution

$$\tan ^{2} \theta=1=\tan ^{2}\left(\frac{\pi}{4}\right) \\$$
$$\tan ^{2} \theta=\tan ^{2} \alpha \Rightarrow \theta=n \pi \pm \alpha \\$$
$$\theta=n \pi \pm \frac{\pi}{4}$$
Question 2
1 / -0

The maximum value of $$\sin x_1 + 2 \sin x_2 + 3\sin x_3$$ is

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution
Question 3
1 / -0

If $$\log_{3} \sin x-\log_{3} \cos x-\log_{3} (1-\tan x)-\log_{3} (1+\tan x)=-1$$, the $$\tan 2x$$ is equal to

A
$$-2$$

B
$$\dfrac{3}{2}$$

C
$$\dfrac{2}{3}$$

D
$$6$$

Solution
Question 4
1 / -0

If $$\sin^{4} \alpha +\cos^{4} \beta +2=4 \sin \alpha . \cos \beta, 0 \le \alpha , \beta \le \dfrac{\pi}{2}$$, then $$(\sin \alpha +\cos \beta)$$ is equal to

A
$$\sqrt{2}$$

B
$$\dfrac{1}{2}$$

C
$$2$$

D
$$1$$

Solution
Question 5
1 / -0

If $$cos \theta=\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)$$ then $$cos3\theta$$ in terms of $$'a'$$=

A
$$\dfrac{1}{4}\left(a^3+\dfrac{1}{a^3}\right)$$

B
$$4\left(a^3+\dfrac{1}{a^3}\right)$$

C
$$\dfrac{1}{2}\left(a^3+\dfrac{1}{a^3}\right)$$

D
None of these

Solution

Given

$$\cos \theta =\dfrac 12\left(a+\dfrac 1a\right)$$

$$\cos 3\theta =4\cos ^3\theta -3\cos \theta $$

$$\implies 4 \left[\dfrac 12\left(a+\dfrac 1a\right)\right]^3-3\left[\dfrac 12\left(a+\dfrac 1a\right)\right]$$

$$\implies \dfrac 48 \left(a+\dfrac 1a\right)^3-\dfrac 32 \left(a+\dfrac 1a\right)$$

$$\implies \dfrac 12 \left( a^3+\dfrac 1{a^3} +3a^2\dfrac 1a+3a\dfrac 1{a^2}\right)-\dfrac 32 \left(a+\dfrac 1a\right)$$

$$\bigg( (a+b)^3=a^3+b^3+3a^2b+3ab^2\bigg)$$

$$\implies \dfrac 12\left( a^3+\dfrac 1{a^3} + 3a + \dfrac 3a \right) -\dfrac 32\left(a-\dfrac 1a \right)$$

$$\implies \dfrac 12\left( a^3+\dfrac 1{a^3} +3\left(a+\dfrac 1a\right) -3\left(a-\dfrac 1a \right)\right)$$

$$\implies \cos3\theta =\dfrac 12 \left( a^3+\dfrac 1{a^3}\right)$$
Question 6
1 / -0

In $$ \Delta ABC, \, \sum \dfrac{a^2 sin (B - C)}{sin A} = $$

A
$$a^2 b^2$$

B
$$b^2 c^2$$

C
$$c^2 a^2$$

D
$$0$$

Solution

In $$\triangle ABC,a=2 R\sin A=2 R\sin (180^{\circ}-(B+C))=2 R\sin (B+C),\dfrac{a}{\sin A}=2 R$$
$$\displaystyle\sum \dfrac{a^2\sin (B-C)}{\sin A}=\displaystyle\sum\dfrac{a}{\sin A}\times a \times \sin(B-C)=\sum \dfrac{a}{\sin A}(2 R\sin (B+C)\sin (B-C))$$
$$=4 R^2\sum (\sin ^2 B-\sin^2 C)$$ $$(\because \sin(A+B)\sin (A-B)=\sin^2 A-\sin^2 B)$$
$$=4 R^2(\sin^2 B-\sin^2 C+\sin^2 C-\sin^2 A+\sin^2 A-\sin^2 B)$$
$$=0$$
Question 7
1 / -0

In $$\Delta ABC, \sum \, a \, sin (B - C) = $$

A
abc

B
2abc

C
3 abc

D
0

Solution

In $$\triangle ABC,a=2 R\sin A=2 R\sin (180^{\circ}-(B+C))=2 R\sin (B+C)$$
$$\displaystyle\sum a\sin (B-C)=2 R\sum \sin (B+C)\sin (B-C)$$
$$=2 R\sum (\sin ^2 B-\sin^2 C)$$ $$(\because \sin(A+B)\sin (A-B)=\sin^2 A-\sin^2 B)$$
$$=2 R(\sin^2 B-\sin^2 C+\sin^2 C-\sin^2 A+\sin^2 A-\sin^2 B)$$
$$=0$$
Question 8
1 / -0

If $$A+B=90,\sin A=a, \sin B=b$$ then $$a^{2}+b^{2}=?$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$None\ of\ these$$

Solution

Given $$A+B=90\implies B=90-A$$
And also given that $$a=\sin A,b=\sin B=\sin (90-A)=\cos A$$
$$a^2+b^2=\sin^2 A+\cos^2 A=1$$
Question 9
1 / -0

If $$\cos A + \cos B = 4 \sin^2 \left(\dfrac{C}{2}\right)$$ then triangle $$ABC$$, which relation holds good

A
$$2b = a + c$$

B
$$b = a + c$$

C
$$a = b + c$$

D
$$2c = a + b$$

Solution
Question 10
1 / -0

In $$\triangle\ {ABC},\left(\cot\dfrac{A}{2}+\cot\dfrac{B}{2}\right)\left(a\sin^{2}\dfrac{B}{2}+b\sin^{2}\dfrac{A}{2}\right)=$$

A
$$\cot{C}$$

B
$$c\cot{C}$$

C
$$\cot\dfrac{C}{2}$$

D
$$c\cot\dfrac{C}{2}$$

Solution