Trigonometric Functions Test 24

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Trigonometric Functions Test 24

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Question 1
1 / -0

In $$\Delta ABC . \sum \dfrac{a^2 sin (B - C)}{sin B + sin C}$$

A
$$a + b $$

B
$$b + c$$

C
$$c + a$$

D
$$0$$

Solution

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

The value of the expression $$\dfrac { 1 - 4 \sin 10 ^ { \circ } \sin 70 ^ { \circ } } { 2 \sin 10 ^ { \circ } }$$ is

A
$$\dfrac { 1 } { 2 }$$

B
$$1$$

C
$$2$$

D
None if these

Solution

$$\dfrac {1-4 sin 10^o sin 70^o}{2 son 10^o}=\dfrac {1-4 sin 10^o sin(90-20)}{2 sin 10^o}$$
$$\dfrac {1-4 sin 10^o cos 20^o}{2 sin 10^o}$$ $$\left \{ since \,\, sin (90-\theta)=cos \theta \right.$$
$$\dfrac {1-2 (sin 10^o cos 20^o)}{2 sin 10^o}$$
$$\dfrac {1-2 sin 30 +sin(-10^o)}{2 sin 10^o}$$ $$\left \{ since\,\, sin(A+B)+Sin (A-B)= 2 Sin A \, Cos B \right.$$
$$\dfrac {1-2(1/2 -sin 10^o)}{2 sin 10^o}$$
$$\dfrac {1-2\times \frac{1}{2} + 2sin 10^o)}{2 sin 10^o}=1$$
Question 3
1 / -0

If $$3 \sin \theta + 5 \cos \theta = 5$$, then the value of $$5 \sin \theta - 3 \cos \theta$$ is equal to

A
$$5$$

B
$$3$$

C
$$4$$

D
None of these

Solution

Let $$5\sin \theta-3\cos \theta=k\cdots(1)$$
Given $$3\sin \theta+5\cos \theta=5\cdots(2)$$
$$(1)^2+(2)^2\implies (5\sin \theta-3\cos \theta)^2+(5\cos \theta+3\sin \theta)^2=k^2+5^2$$
$$\implies 25\sin^2 \theta+9\cos ^2 \theta-30\sin \theta\cos \theta+25\cos^2 \theta+9\sin ^2 \theta+30\sin \theta\cos \theta=k^2+25$$
$$\implies 25(\sin^2 \theta+\cos^2 \theta )+9(\sin^2 \theta+\cos^2 \theta)=k^2+25$$
$$\implies k^2=25+9-25=9\implies k=\pm3$$
So $$5\sin \theta-3\cos \theta=3$$
Question 4
1 / -0

The general value of satisfying both $$\sin \theta = \dfrac{-1}{2}$$ and $$\tan \theta = \dfrac{1}{\sqrt{3}}$$ is:

A
$$2 n\pi, n \in Z$$

B
$$2n\pi + \dfrac{7\pi}{6 },n\in Z$$

C
$$n\pi + \dfrac{\pi}{4},n \in Z$$

D
$$2 n \pi + \dfrac{\pi}{4},n \in Z$$

Solution
Question 5
1 / -0

$$\displaystyle \sum _{ \lambda =1 }^{ 10 }{ \sin ^{ }{ \left( \sin { \left( \lambda \pi -\dfrac { \pi }{ 6 } \right) } \right) } } $$ is equal to

A
$$\dfrac{5\pi}{3}$$

B
$$\dfrac{\pi}{3}$$

C
$$0$$

D
$$5$$

Solution
Question 6
1 / -0

The minimum value of $$3 \tan^2 \theta + 12 \cos^2 \theta$$:

A
$$6$$

B
$$8$$

C
$$10$$

D
None of these

Solution
Question 7
1 / -0

$${ cos }^{ 2 }\frac { 3\pi }{ 5 } +{ cos }^{ 2 }\frac { 4\pi }{ 5 } =$$ ____________ .

A
$$\frac { 4 }{ 5 } $$

B
$$\frac { 5 }{ 2 } $$

C
$$\frac { 5 }{ 4 } $$

D
$$\frac { 3 }{ 4 } $$

Solution
Question 8
1 / -0

Number of solutions to the equation $$2 ^ { \sin ^ { 2 } x } + 5.2 ^ { \cos ^ { 2 } x } = 7 ,$$ in the interval $$[ - \pi , \pi ] ,$$ is

A
$$2$$

B
$$4$$

C
$$6$$

D
none of these

Solution

$$2^{\sin^2x}+5.2^{\cos^2x}=7$$

$$2^{\sin^2x}+5.2^{1-\sin^2x}=7$$

Let $$2^{\sin^2x}=k$$

$$k+5\cdot\dfrac{2}{k}=7$$

$$k+\dfrac{10}{k}=7$$

$$k^2+10=7h$$

$$k^2-7k+10=0$$

$$(k-2)(k-5)=0$$

But $$2^{\sin^2x}\leq 2'=2$$

Hence $$2^{\sin^2x}\neq 5$$

$$\sin^2x=1$$

$$\Rightarrow x=\dfrac{\pi}{2}, \dfrac{-\pi}{2}$$

Hence there are only two solution in $$[-\pi, \pi]$$ i.e., $$\dfrac{-\pi}{2}$$ and $$\dfrac{\pi}{2}$$.
Question 9
1 / -0

If $$5\left({\tan}^{2}x-{\cos}^{2}x\right)=2\cos2x+9$$, then the value of $$\cos4x$$ is:

A
$$-\dfrac{3}{5}$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{2}{9}$$

D
$$-\dfrac{7}{9}$$

Solution
Question 10
1 / -0

$$cos \theta - 4 sin \theta = 1\Rightarrow sin\theta +4cos\theta $$

A
$$\pm 1$$

B
0

C
$$\pm 2$$

D
$$\pm 4$$

Solution

Let $$\sin \theta+4\cos \theta=k\cdots(1)$$
Given $$\cos \theta-4\sin \theta=1\cdots(2)$$
$$(1)^2+(2)^2\implies (\sin \theta+4\cos \theta)^2+(\cos \theta-4\sin \theta)^2=k^2+1^2$$
$$\implies \sin^2 \theta+16\cos ^2 \theta+8\sin \theta\cos \theta+\cos^2 \theta+16\sin ^2 \theta-8\sin \theta\cos \theta=k^2+1$$
$$\implies (\sin^2 \theta+\cos^2 \theta )+16(\sin^2 \theta+\cos^2 \theta)=k^2+1$$
$$\implies k^2=1+16-1=16\implies k=\pm 4$$
So $$\sin \theta+4\cos \theta=\pm 4$$