Trigonometric Functions Test 34

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Trigonometric Functions Test 34

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Weekly Quiz Competition

Question 1
1 / -0

If $$a+b=3-\cos { 4\theta } $$ and $$a-b=4\sin { 2\theta } $$, then (ab) is always less then equal to

A
1/2

B
1

C
2/3

D
3/4

Solution
Question 2
1 / -0

If $$a=\cos { \alpha +\cos { \beta -\cos { (\alpha +\beta ),\quad b=4\sin { \frac { \alpha }{ 2 } \sin { \frac { \beta }{ 2 } \cos { (\frac { \alpha +\beta }{ 2 } ) } } } } } } $$ then a-b=

A
0

B
1

C
-1

D
-2

Solution
Question 3
1 / -0

If $$\csc\theta=\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}$$ where $$x$$ and $$y$$ are two unequal non-zero real numbers, then the number of real values of $$\theta$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$infinite$$

Solution
Question 4
1 / -0

$$\displaystyle\sum_{r=1}^{n-1} \cos^2\left ( \dfrac{r\pi}{n} \right )=? $$

A
$$\dfrac{n}{2}$$

B
$$\dfrac{n-2}{2}$$

C
$$\dfrac{n-1}{2}$$

D
$$\dfrac{n-2}{4}$$

Solution

As we know that
$$\cos 2 A=2 \cos ^{2} A-1$$
$$\cos ^{2} A=\frac{1+\cos 2 A}{2}$$
say , $$A=\frac{\pi \pi}{n}$$
$$\Rightarrow \sum_{r=1}^{n-1} 1+\cos 2\left(\frac{r \pi}{n}\right)=\frac{1}{2}\left[\sum_{r=1}^{n-1} \cos \frac{2 \pi \pi}{n}+\sum_{n=1}^{n-1} 1\right]$$
$$\Rightarrow \frac{1}{2}\left[\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\ldots+\cos \frac{2(n-1)}{n} \pi+(n-1)\right]$$

As we know that
$$\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\cdots+\cos (\alpha+(n-1) \beta)=\frac{\sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}} \cdot \cos \left(\frac{2 \alpha+(n-1) \beta}{2}\right) \\$$
$$= \frac{1}{2}\left[\frac{\sin \left(\frac{(n-1) \pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)} \cdot \cos \left(\frac{\frac{4\pi}{n}+(n-2) \frac{2 \pi}{n}}{2}\right)+n-1\right]$$

$$= \frac{1}{2}\left[\frac{-\sin \left(\frac{n-1}{n} \pi\right)}{\sin \frac{\pi}{n}}+n-1\right]$$

$$\left.= \frac{1}{2}\left[-\left\{\sin \frac{\sin \left(\pi-\frac{(n-1)}{n}\right) \pi}{\sin \frac{\pi}{n}}\right\}\right\}+n-1\right]$$

$$= \frac{1}{2}[-\frac{sin\frac{\pi}{n}}{sin\frac{\pi}{n}}+n-1]$$

$$= \frac{1}{2}(-1+n-1)=\frac{n-2}{2}$$
Hence, $$(B)$$ is the correct option.
Question 5
1 / -0

Consider the curve $$x=1-{ 3t }^{ 2 },y=t-{ 3t }^{ 3 }$$.
If a tangent at point $$(1-{ 3t }^{ 2 },t-{ 3t }^{ 3 })$$ inclined at an angel $$\theta $$ to the positive x-axis and another tangent at point P(-2,2) cuts the curve angels at Q.
The value of $$tan\theta +sec\theta $$ is equal to

A
3t

B
t

C
$$t-{ t }^{ 2 }$$

D
$${ t }^{ 2 }-2t$$

Solution
Question 6
1 / -0

If $$\cos \theta = \dfrac{\cos \alpha - \cos \beta}{1 - \cos \alpha \cdot \cos \beta}$$ then $$\tan \left(\dfrac{\theta}{2}\right)$$ is _____

A
$$\cot \dfrac{\beta}{2} \tan \dfrac{\alpha}{2}$$

B
$$\tan \alpha \tan \dfrac{\beta}{2}$$

C
$$\tan \dfrac{\beta}{2} \cot \dfrac{\alpha}{2}$$

D
$$\tan^2 \dfrac{\alpha}{2} \tan^2 \dfrac{\beta}{2}$$

Solution
Question 7
1 / -0

The value of $$\sum _{ r=1 }^{ 10 }{ \cos ^{ 3 }{ \frac { r\pi }{ 3 } } } =$$?

A
$$-\frac { 1 }{ 8 } $$

B
$$-\frac { 7 }{ 8 } $$

C
$$-\frac { 9 }{ 8 } $$

D
None of the above

Solution
Question 8
1 / -0

$$\dfrac { \tan\theta }{ 1-\cot\theta } +\dfrac { \cot\theta }{ 1-\tan\theta } $$ is equal to

A
$$1+\tan\theta +\cot\theta $$

B
$$1-\tan\theta -\cot\theta $$

C
$$1+\tan\theta -\cot\theta $$

D
none of these

Solution
Question 9
1 / -0

For $$-\frac { \pi }{ 2 } <\theta <\frac { \pi }{ 2 } ,\frac { \sin { \theta } +\sin { 2\theta } }{ 1+\cos { \theta } +\cos { 2\theta } } $$ lies in the interval

A
$$\left( -\infty ,\infty \right) $$

B
$$(-2,2)$$

C
$$\left( 0,\infty \right) $$

D
$$(-1,1)$$

Solution
Question 10
1 / -0

$$ \operatorname{Tan}\left[\frac{1}{2} \sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)\right]= $$

A
$$

\frac{2 a}{1+a^{2}}

$$

B
$$

\frac{2 a}{1-a^{2}}

$$

C
$$

\frac{a}{1+a^{2}}

$$

D
$$

\frac{a}{1-a^{2}}

$$