Introduction to Trigonometry Test

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Introduction to Trigonometry Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

Given $$a = \sec^{2} \alpha ,b=\mathrm{cosec}^{2} \alpha , and , c=\dfrac{1}{1-\sin^{2}\alpha \cos^{2}\alpha }.$$ The value of $$bc + ca - ab$$ is

A
$$a$$

B
$$b$$

C
$$c$$

D
$$-c$$

Solution

We know that,
$${ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1$$

Given that, $$a = \sec^{2} \alpha ,b=\mathrm{cosec}^{2} \alpha , $$

$$\Rightarrow$$ $$\cfrac { 1 }{ a } ={ cos }^{ 2 }\alpha ,\quad \cfrac { 1 }{ b } ={ sin }^{ 2 }\alpha $$

$$\dfrac{1}{a}+\dfrac{1}{b}=1$$

and $$c =\dfrac{1}{1-\dfrac{1}{ab}}=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{ab}}=\dfrac{ab}{a+b-1}$$

$$\therefore ac + bc- c = ab$$,

Thus,
$$ac + bc - ab = c$$

Hence, option 'C' is correct.
Question 2
1 / -0

lf $$\sin\alpha=a\sin\beta$$ and $$\cos\alpha=b\cos\beta$$ , then $$\tan\alpha$$=

A
$$\pm\sqrt{\displaystyle \dfrac{a^{2}(1-b^{2})}{b^{2}(a^{2}-1)}}$$

B
$$\pm\sqrt{\displaystyle \dfrac{1-b^{2}}{a^{2}-1}}$$

C
$$\pm\sqrt{\displaystyle \dfrac{1+b^{2}}{a^{2}-1}}$$

D
$$none\ of\ these$$

Solution

Given that $$\sin\alpha=a\sin\beta$$ and $$\cos\alpha=b\cos\beta$$
$$\Rightarrow \sin^2\alpha=a^2\left(1-\cos^2\beta\right)$$
$$\Rightarrow \sin^2\alpha=a^2\left(1-\displaystyle\frac{1}{b^2}\cos^2\alpha\right)$$
dividing both sides with $$\cos^2\alpha$$ gives
$$\tan^2\alpha=a^2\left(1+\tan^2\alpha\right)-\displaystyle\frac{a^2}{b^2}$$
$$\Rightarrow \tan^2\alpha(1-a^2)=\displaystyle\frac{a^2}{b^2}(b^2-1)$$
$$\therefore \tan\alpha=\pm\sqrt{\displaystyle \frac{a^{2}(1-b^{2})}{b^{2}(a^{2}-1)}}$$
Hence, option A.
Question 3
1 / -0

If $$\sin\mathrm{x}+\sin^{2}\mathrm{x}=1$$, then $$\cos^{12}\mathrm{x}+3\cos^{10}\mathrm{x}+3\cos^{8}\mathrm{x}+\cos^{6}\mathrm{x}+2\cos^{4}\mathrm{x}+\cos^{2}\mathrm{x}-2$$ is equal to

A
$$0$$

B
$$\cos^{2}\mathrm{x}$$

C
$$\sin^{2}\mathrm{x}$$

D
$$-\sin^{2}\mathrm{x}$$

Solution

Given that,
$$\sin x + \sin^{2}x = 1$$

To find out,
The value of the expression: $$\cos^{12} x+3\cos^{10}x+3\cos^{8}x+\cos^{6}x+2\cos^{4}x+\cos^{2}x-2$$

$$\sin x + \sin^{2}x = 1$$

$$\Rightarrow \sin x = 1 - \sin^{2}x$$

$$\Rightarrow \sin x = \cos^{2}x\quad \quad [\because \ \sin^2 x+\cos^2 x=1]$$

$$\cos^{12} x+3\cos^{10}x+3\cos^{8}x+\cos^{6}x+2\cos^{4}x+\cos^{2}x-2$$

We can rewrite it as:
$$({\cos^2x})^6+3({\cos^2x})^5+3({\cos^2x})^4+({\cos^2x})^3+2({\cos^2x})^2+{\cos^2x}-2$$

Now, substituting $$\cos^{2}x = \sin x$$ in the above expression, we get:

$$ \sin^{6}x + 3\sin^{5}x + 3\sin^{4}x + \sin^{3}x + 2\sin^{2}x + \sin x -2$$

$$\Rightarrow \left[(\sin^2x)^3+(\sin x)^3+3\sin x \sin^2 x(\sin^2 x+\sin x)\right]-2(1-\sin^2 x)+\sin x$$

We know that, $$a^3+b^3+3ab(a+b)=(a+b)^3$$

On applying the identity, the expression becomes,
$$(\sin^2 x+\sin x)^3-2(1-\sin^2 x)+\sin x$$

Now, $$\sin x + \sin^{2}x = 1$$ and $$1-\sin^2 x=\sin x$$

Substituting the above values, the expression becomes,
$$(1)^3-2(\sin x)+\sin x$$

$$\Rightarrow 1-2\sin x+\sin x$$

$$\Rightarrow 1-\sin x$$

But, $$1-\sin x=\sin^2 x\quad \quad [\because \ \sin x + \sin^{2}x = 1]$$

Hence, the expression reduces to,
$$\sin^2 x$$

$$\therefore \ \cos^{12} x+3\cos^{10}x+3\cos^{8}x+\cos^{6}x+2\cos^{4}x+\cos^{2}x-2=\sin^2 x$$

Hence, option C is correct.
Question 4
1 / -0

If $$\sin {x}+\sin^{2}{x}+\sin^{3}{x}=1 ,\ \ then \ \ \cos^{6}{x}-4\cos^{4}{x}
+8\cos^{2}{x}$$ is equal to

A
$$4$$

B
$$2$$

C
$$1$$

D
$$0$$

Solution

$$\sin x + \sin ^{2}x + \sin ^{3}x = 1$$
$$\sin x(1 + \sin ^{2}x) = 1 - \sin ^{2}x$$
$$\sin x(2 - \cos ^{2}x) = \cos ^{2}x$$
we square both the sides
$$\sin ^{2}x(2-\cos ^{2}x)^{2} = \cos ^{4}x$$
$$(1-\cos ^{2}x)(2-\cos ^{2}x)^{2} = \cos ^{4}x$$
$$(1-\cos^2x)(4-4\cos^2 x+\cos^4x)=\cos ^4 x$$
$$8\cos^2x+\cos^6x-4\cos^4x=4$$
Question 5
1 / -0

If $$\sin\alpha+\cos\alpha=m, m^{2}\leq 2$$, then $$\sin^{6}\alpha+\cos^{6}\alpha=$$

A
$$\displaystyle \frac{4+3(m^{2}+1)^{2}}{4}$$

B
$$\displaystyle \frac{4-3(m^{2}+1)^{2}}{4}$$

C
$$\displaystyle \frac{4-3(m^{2}-1)^{2}}{4}$$

D
$$\displaystyle \frac{4+3(m^{2}-1)^{2}}{4}$$

Solution

$$\sin\alpha + \cos\alpha = m$$
squaring both sides
$$\sin^{2}\alpha + \cos^{2}\alpha + 2\sin\alpha \cos\alpha = m^{2}$$

$$\sin\alpha \cos\alpha = \dfrac{m^{2} - 1}{2}$$
Now,
$$ sin^{6}\alpha + cos^{6}\alpha = (sin^{2}\alpha + cos^{2}\alpha)(\sin^{4}\alpha + cos^{4}\alpha - sin^{2}\alpha cos^{2}\alpha)$$

$$ = (sin^{2}\alpha + \cos^{2}\alpha)^{2} - 3 \sin^{2}\alpha \cos^{2}\alpha$$

$$ = 1 - 3\sin^{2}\alpha \cos^{2}\alpha$$

Substitute the value of $$\sin\alpha \cos\alpha$$

$$=1-3\dfrac{(m^2-1)^2}{4}$$
Question 6
1 / -0

If $$a \sin^{2}\theta+b\cos^{2}\theta=a\cos^{2}\phi+b\sin^{2}\phi=1$$ and $$a \tan\theta=b\tan\phi$$, then choose the correct option.

A
$${a+b}=2ab$$

B
$${a-b}=2ab$$

C
$${a-b}+2ab=0$$

D
$${a+b}+2ab=0$$

Solution

Given: $$a\sin ^2\theta+b\cos ^2\theta=1$$ .... $$(i)$$ and $$a\cos ^2\phi+b\sin ^2\phi=1$$ .... $$(ii)$$
Dividing equation $$(i)$$ by $$\cos ^2\theta$$, we get

$$a\tan ^2\theta+b=\sec ^2\theta$$
$$\Rightarrow a\tan ^2\theta+b=1+\tan ^2\theta$$
$$\Rightarrow \tan ^2\theta=\dfrac{b-1}{1-a}=\tan \theta=\sqrt{\dfrac{b-1}{1-a}}$$

Similarly in the equation $$(ii)$$, on dividing by $$\cos ^2\phi$$ we get,
$$b\tan ^2\phi+a=\sec ^2\phi$$
$$\Rightarrow b\tan ^2\phi+a=1+\tan ^2\phi$$
$$\Rightarrow \tan ^2\phi=\dfrac{1-a}{b-1}=\tan \phi=\sqrt{\dfrac{1-a}{b-1}}$$

Now, it is given that
$$a\tan \theta=b\tan \phi$$
$$\Rightarrow a\sqrt{\dfrac{b-1}{1-a}}=b\sqrt{\dfrac{1-a}{b-1}}$$
$$\Rightarrow \dfrac{a}{b}=\dfrac{1-a}{b-1}$$
$$\Rightarrow ab-a=b-ab$$
$$\Rightarrow a+b=2ab$$
Question 7
1 / -0

The value of the expression $$(\tan1^{0} \tan2^{0} \tan 3^{0}...\tan89^{0})$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$\dfrac{1}{2}$$

Solution

Given:
$$A = \tan1 ^o\cdot\tan2^o\cdot\tan3^o\cdots\cdots\tan89^o$$
We know that, the value of $$\tan89^o=\tan(90^o-1^o)=\cot 1^o$$
In the same way,
$$\tan 88^o=\cot 2^o$$
$$\tan 87^o=\cot 3^o$$
And .... so on, up to $$\tan 46^o=\cot 44^o$$
And the middle one is $$\tan 45^o=1$$
So,
$$A = \tan1^o\cdot\tan2^o\cdot\tan3^o\cdots\cdots\tan{44}^o\cdot\tan45^o\cdot\cot44^o\cdots\cdots\cot 2^o\cdot\cdot\cot1^o$$
$$\tan$$ and $$\cot$$ cancels out each other, then $$\tan45^o = 1$$ is remaining
$$\therefore$$ The value of the expression $$V$$ is $$1$$.

Hence, option B.
Question 8
1 / -0

If $$b > 1, \sin t > 0, \cos t > 0$$ and $$\displaystyle \log_b(\sin t) = x$$, then $$\displaystyle \log_b(\cos t)$$ is equal to

A
$$\displaystyle \frac {1}{2} \log_b(1 - b^{2x})$$

B
$$\displaystyle 2\log (1 - b^{x/2})$$

C
$$\displaystyle \log_b \sqrt {1 - b^{2x}} \log_b(1 - b^{2x})$$

D
$$\displaystyle \sqrt {1 - x^2}$$

Solution

Hence
$$sint=b^{x}$$
Or
$$sin^{2}t=b^{2x}$$
$$1-cos^{2}t=b^{2x}$$
$$cost=\sqrt{1-b^{2x}}$$
Or
$$log_{b}(cost)=\dfrac{1}{2}log_{b}(1-b^{2x})$$.
Question 9
1 / -0

If $$\cos9 \alpha= \sin \alpha$$ and $$9 \alpha < 90^{0}$$, then the value of $$\tan5 \alpha$$ is

A
$$\dfrac{1}{\sqrt{3}}$$

B
$$\sqrt{3}$$

C
$$1$$

D
$$0$$

Solution

$$\cos(9\alpha)=\sin(\alpha)$$
$$\sin(90^0-9\alpha)=\sin(\alpha)$$

For $$0<\alpha<90^0$$
$$90^0-9\alpha=\alpha$$
$$90^0=10\alpha$$
$$\alpha=9^0$$
Now,
$$\tan5\alpha$$
$$=\tan(5)(9^0)$$
$$=\tan45^0 = 1$$

Hence, option C.
Question 10
1 / -0

In a right angle triangle $$\triangle ABC,\,\sin ^{ 2 }{ A } +\sin ^{ 2 }{ B } +\sin ^{ 2 }{ C } $$ is

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of these

Solution

Let $$C={90}^{0}$$
Then,
$$\sin ^{ 2 }{ A } +\sin ^{ 2 }{ B } +\sin ^{ 2 }{ C } =\sin ^{ 2 }{ A } +\sin ^{ 2 }{ B } +1$$
$$\displaystyle =\sin ^{ 2 }{ A } +\sin ^{ 2 }{ \left( \frac { \pi }{ 2 } -A \right) +1 } \\ =\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } +1\\=1+1 \\=2$$

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Introduction to Trigonometry Test - 68

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Introduction to Trigonometry Test - 68

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SHARING IS CARING

Question 1

$$a$$

$$b$$

$$c$$

$$-c$$

Solution

Question 2

$$\pm\sqrt{\displaystyle \dfrac{a^{2}(1-b^{2})}{b^{2}(a^{2}-1)}}$$

$$\pm\sqrt{\displaystyle \dfrac{1-b^{2}}{a^{2}-1}}$$

$$\pm\sqrt{\displaystyle \dfrac{1+b^{2}}{a^{2}-1}}$$

$$none\ of\ these$$

Solution

Question 3

$$0$$

$$\cos^{2}\mathrm{x}$$

$$\sin^{2}\mathrm{x}$$

$$-\sin^{2}\mathrm{x}$$

Solution

Question 4

$$4$$

$$2$$

$$1$$

$$0$$

Solution

Question 5

$$\displaystyle \frac{4+3(m^{2}+1)^{2}}{4}$$

$$\displaystyle \frac{4-3(m^{2}+1)^{2}}{4}$$

$$\displaystyle \frac{4-3(m^{2}-1)^{2}}{4}$$

$$\displaystyle \frac{4+3(m^{2}-1)^{2}}{4}$$

Solution

Question 6

$${a+b}=2ab$$

$${a-b}=2ab$$

$${a-b}+2ab=0$$

$${a+b}+2ab=0$$

Solution

Question 7

$$0$$

$$1$$

$$2$$

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

Solution

Question 8

$$\displaystyle \frac {1}{2} \log_b(1 - b^{2x})$$

$$\displaystyle 2\log (1 - b^{x/2})$$

$$\displaystyle \log_b \sqrt {1 - b^{2x}} \log_b(1 - b^{2x})$$

$$\displaystyle \sqrt {1 - x^2}$$

Solution

Question 9

$$\dfrac{1}{\sqrt{3}}$$

$$\sqrt{3}$$

$$1$$

$$0$$

Solution

Question 10

$$0$$

$$1$$

$$-1$$

None of these

Solution

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