CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 4 - MCQExams.com

Given $$\sin A = \dfrac{3}{5}$$
$$\sin A =\dfrac{P}{B} = \dfrac{3}{5}$$
Thus, $$P = 3, H = 5$$

By Pythagoras Theorem,
$$H^2 = P^2 + B^2$$
$$5^2 = 3^2 + B^2$$
$$B= 4 $$ cm

Now, $$\tan A = \dfrac{P}{B} = \dfrac{3}{4}$$

$$\cos { \left( p\sin { x }  \right)  } =\sin { \left( p\cos { x }  \right)  } $$
$$\displaystyle \Rightarrow \cos { \left( p\sin { x }  \right)  } =\cos { \left( \frac { \pi  }{ 2 } -p\cos { x }  \right)  } $$

$$\displaystyle \Rightarrow p\sin { x } =2n\pi \pm \left( \frac { \pi  }{ 2 } -p\cos { x }  \right) $$

$$\displaystyle \Rightarrow p\sin { x } +p\cos { x } =2n\pi +\frac { \pi  }{ 2 } $$ or $$\displaystyle p\sin { x } -p\cos { x } =2n\pi -\frac { \pi  }{ 2 } $$

$$\displaystyle \Rightarrow p\sqrt { 2 } \left( \sin { \left( x+\frac { \pi  }{ 4 }  \right)  }  \right) =\frac { \left( 4n+1 \right) \pi  }{ 2 } $$
Or $$\displaystyle p\sqrt { 2 } \left( \sin { \left( x-\frac { \pi  }{ 4 }  \right)  }  \right) =\frac { \left( 4n-1 \right) \pi  }{ 2 } \quad $$

As $$\displaystyle -1\le \sin { \left( x+\frac { \pi  }{ 4 }  \right)  } \le 1$$
$$\displaystyle \Rightarrow -p\sqrt { 2 } \le p\sqrt { 2 } \sin { \left( x+\frac { \pi  }{ 4 }  \right)  } \le p\sqrt { 2 } $$
$$\displaystyle \Rightarrow -p\sqrt { 2 } \le \frac { \left( 4n+1 \right) \pi  }{ 2 } \le p\sqrt { 2 } $$   ...(1)

And as $$\displaystyle -1\le \sin { \left( x-\frac { \pi  }{ 4 }  \right)  } \le 1$$
$$\displaystyle \Rightarrow -p\sqrt { 2 } \le p\sqrt { 2 } \sin { \left( x-\frac { \pi  }{ 4 }  \right)  } \le p\sqrt { 2 } $$
$$\displaystyle \Rightarrow -p\sqrt { 2 } \le \frac { \left( 4n-1 \right) \pi  }{ 2 } \le p\sqrt { 2 } $$   ...(2)
(2) is always a subset of of first, therefore we have to consider only first 
It is sufficient to consider $$n\ge 0$$ , because for $$n>0$$, the solution will be same for $$n\ge 0$$

If $$n\ge 0$$ $$\displaystyle -\sqrt { 2 } p\le \frac { \left( 4n+1 \right) \pi  }{ 2 } \Rightarrow \frac { \left( 4n+1 \right) \pi  }{ 2 } \le \sqrt { 2 } p$$

$$\displaystyle \sqrt { 2 } p\ge \frac { \pi  }{ 2 } \Rightarrow p\ge \frac { \pi  }{ 2\sqrt { 2 }  } =\frac { \pi \sqrt { 2 }  }{ 4 } $$

$$ tanx + secx = 2  cosx$$
$$\Rightarrow sinx + 1 = 2  cos^2x \Rightarrow sinx+1=2 - 2  sin^2x$$
$$ \Rightarrow 2  sin^2x + sinx -1 = 0$$
$$ \Rightarrow (2  sinx - 1) (sinx + 1) = 0$$
but $$ sinx = -1\Rightarrow x=\dfrac{3\pi}{2}$$ ...........(1)
$$ sinx = \dfrac{1}{2} = sin\dfrac{\pi}{6}$$
therefore general solution is,
$$ x = n\pi + (-1)^n.\dfrac{\pi}{6}$$
$$ x = ......, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, ......$$     ..........(2)
Therefore, number of solutions in the given interval are $$3$$.
Hence, option 'D' is correct.

Given that $$\tan A = 1/2,\quad \tan B = 1/3$$

$$\tan(A+B) = \displaystyle\frac{\tan A + \tan B}{1-\tan A\tan B}$$

$$\quad = \displaystyle\frac{\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}}{1-\left(\displaystyle\frac{1}{2}\right)\left(\displaystyle\frac{1}{3}\right)} = \displaystyle\frac{5}{6}\times\displaystyle\frac{6}{5} = 1$$

$$\quad \therefore \tan(A+B) = 1 = \tan45^{\small\circ} \quad \Rightarrow A+B = 45^{\small\circ}$$

$$=cos(60^{0}-30^{0})$$
$$=cos(30^{0})$$
$$=\frac{\sqrt{3}}{2}$$
$$=LHS$$
Now 
$$RHS$$
$$=cos60^{0}cos30^{0}+sin30^{0}sin60^{0}$$
$$=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$$
$$=\frac{\sqrt{3}}{2}$$
$$=RHS$$
Hence verified.

$$=cot(60^{0}+30^{0})$$
$$=cot(90^{0})$$
$$=0$$
$$=LHS$$
RHS
$$=\frac{cot60^{0}.cot30^{0}-1}{cot60^{0}+cot30^{0}}$$
$$=\dfrac{1-1}{\frac{1}{\sqrt{3}}+\sqrt{3}}$$
$$=0$$
$$Hence\ verified.$$

Simplifying the given expression, we get
$$\dfrac{cosec\theta-1+cosec\theta+1}{\sqrt{cosec^{2}\theta-1}}$$

We know that $$\tan 2A$$ $$=\dfrac{2\tan A}{1-\tan^{2}A}$$
Here $$A=30^{0}$$
Substituting, we get
$$\dfrac{2\tan30^{0}}{1-\tan^{2}30^{0}}$$
$$=\tan(2(30^{0}))$$
$$=\tan(60^{0})$$
$$=\sqrt{3}$$

$$\dfrac{cos(\theta)(1-sin\theta+1+sin\theta)}{1-sin^{2}\theta}$$
$$=\dfrac{cos(\theta)(2)}{cos^{2}\theta}$$
$$=\dfrac{2}{cos\theta}$$
$$=2sec\theta$$

It is given that
$$\alpha=2\beta$$
Hence
$$\alpha+\beta=90^{0}$$
Substituting, we get
$$3\beta=90^{0}$$
Hence
$$\beta=30^{0}$$
Therefore $$\alpha=60^{0}$$
Hence
$$\sin^{2}\beta+\cos^{2}\alpha$$
$$=\dfrac{1}{2}^{2}+\dfrac{1}{2}^{2}$$
$$=\dfrac{1}{4}+\dfrac{1}{4}$$
$$=\dfrac{1}{2}$$

Given: $$p \sin x = q$$
Now, $$\displaystyle \sqrt{p^{2}-q^{2}}\tan x$$
$$=\sqrt{p^{2}-p^{2}\sin ^{2}x.}\tan \:x$$
$$=\sqrt{p^{2}\left ( 1-\sin ^{2}x \right )}.\tan \:x$$
$$=p\sqrt{\left ( 1-\sin ^{2}x \right )}.\tan \:x\left .....( \because \cos ^{2}\theta +\sin ^{2}\theta =1 \right )$$
$$=P\sqrt{\cos ^{2}x}.\tan \:x$$
$$=p\cos \:x\tan \:x$$
$$=p\sin \:x$$
$$=q$$

When, $$x \in [0, \frac{\pi}{2}]$$ and $$x \in [\pi, \frac{3\pi}{2}]$$
Thus, $$\tan x > 0$$
hence, $$|\tan x | = \tan x$$
hence, $$\displaystyle  \left | \tan x \right |=\tan x+\frac{1}{\cos x}(0\leq  \times \leq 2\pi)$$ 
= $$\displaystyle  \tan x =\tan x+\frac{1}{\cos x}$$
= $$\cos x = 1$$
= $$\cos x = \cos 90$$
= $$x = 90^{\circ}$$
But it is not possible since, $$\cos x \ne 0$$ ($$\cos x$$  being the denominator in the equation)

When, $$x \in [\frac{\pi}{2}, \pi ]$$ and $$x \in [\frac{3\pi}{2}, 2\pi]$$
Thus, $$\tan x < 0$$
hence, $$|\tan x | = - \tan x$$
hence, $$\displaystyle  \left | \tan x \right |=\tan x+\frac{1}{\cos x}$$ 
= $$2 \tan x = \frac{1}{\cos x}$$
= $$\sin x = \frac{1}{2}$$
This is not possible, as $$\sin x$$ is not positive in these intervals of value of $$x$$

Hence, the equation has no solution for any value of x.

$$\displaystyle 8\tan A=15 \Rightarrow \tan A=\frac{15}{8}$$

Now, $$\dfrac{\sin A-\cos A}{\sin A+\cos A}$$
Multiply and divide by $$\cos A$$
$$=\dfrac{{\dfrac{\sin A}{\cos A}}-\dfrac{\cos A}{\cos A}}{{\dfrac{\sin A}{\cos A}}+\dfrac{\cos A}{\cos A}}$$

$$\displaystyle\frac { \sec { x }  }{ 1-\cos { x }  } =\frac { 1 }{ 1-\cos { x }  } $$
For $$1-\cos { x } \neq 0\Rightarrow \cos { x } \neq 1$$   ..(1)
Gives $$\sec { x } =1\Rightarrow \cos { x } =1$$   ...(2)
From (1) and (2) we get no solution.

Given, $$x=2{ \sin }^{ 2 }\theta ,y=2{ \cos }^{ 2 }\theta +1$$

$$(5+4\cos\,\theta)\;(2\cos\,\theta+1)=0$$
$$\Rightarrow \cos\,\theta=-\displaystyle\frac{5}{4}$$ (not possible) $$\cos\,\theta=-\displaystyle\frac{1}{2}$$
$$\Rightarrow \cos\theta=-\displaystyle\frac{1}{2}$$
$$\Rightarrow \cos \theta=-\cos {\dfrac{\pi}{3}}=\cos(\pi\pm\dfrac{\pi}{3})$$
$$\Rightarrow \theta=\displaystyle\frac{2\pi}{3},\displaystyle\frac{4\pi}{3}$$   (II and III quad, cosine is negaitve)

$$\sin A (1 + \tan A) + \cos A(1 + \cot A)$$
= $$\sin A(1 + \frac{\sin A}{\cos A}) + \cos A (1 + \frac{\sin A}{\cos A})$$
= $$(\sin A + \cos A) (\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A})$$
= $$(\sin A + \cos A) (\frac{\sin^2 A + \cos^2 A}{\sin A + \cos A}$$
= $$\sec A + cosec A$$
Thus, 1 satisfies the given equation.

Given, $$\sin x + \sin^2 x = 1$$

Given, $$\cos\theta=\dfrac{5}{13}$$
Then $$\sin\theta=\dfrac{12}{13}$$
Hence, $$\tan\theta=\dfrac{12}{5}$$ and
$$\text {cosec} \theta=\dfrac{13}{12}$$
Substituting in the equation, we get 
$$=\dfrac{\frac{5}{13}+\frac{5.5}{12}}{\frac{13}{12}-\frac{5}{13}}$$
$$=\dfrac{5(12)+25(13)}{13(13)-5(12)}$$
$$=\dfrac{60+300+25}{169-60}$$
$$=\dfrac{385}{109}$$

$$\displaystyle \frac{\tan ^{2}\theta }{1+\sec \theta }+1=\frac{\sec^2\theta-1 }{1+\sec \theta }+1$$

The value of $$\displaystyle \dfrac  {(1+\tan ^{2}\theta )}{(1+\cot ^{2}\theta )}$$ is
$$=\displaystyle \dfrac {\left (1+\frac {\sin ^{2}\theta}{\cos ^{2}\theta} \right)}{\left (1+\frac {\cos ^{2}\theta}{\sin ^{2}\theta} \right)}$$
$$=\displaystyle \dfrac {\left (\frac {\cos ^{2}\theta + \sin ^{2}\theta}{\cos ^{2}\theta} \right)}{\left (\frac {\sin ^{2}\theta + \cos ^{2}\theta}{\sin ^{2}\theta}\right )}$$
$$=\displaystyle \dfrac  {\left (\frac {1}{\cos ^{2}\theta}\right )}{\left (\frac {1}{\sin ^{2}\theta} \right)}$$
$$=\displaystyle \frac {\sin ^{2}\theta}{\cos ^{2}\theta}$$
$$=\displaystyle \tan^{2}\theta $$
Hence, option A is correct.

$$\displaystyle \sqrt{\dfrac{1+\cos  A}{1-\cos  A}}=\sqrt{\dfrac{1+\cos  A}{1-\cos  A}\times\dfrac{1+\cos  A}{1+\cos  A}}$$

 $$\displaystyle \left(1-\tan A+\sec A\right)\left(1-\cot A+\text{cosec }  A\right) $$

$$h=\sqrt { 5 } x$$

Given, 

$$(1+\tan {\theta}+\sec {\theta})(1+\cot {\theta}-co\sec {\theta})$$
$$\left( 1+\cfrac { \sin { \theta  }  }{ \cos { \theta  }  } +\cfrac { 1 }{ \cos { \theta  }  }  \right) \left( 1+\cfrac { \cos { \theta  }  }{ \sin { \theta  }  } -\cfrac { 1 }{ \sin { \theta  }  }  \right) $$
$$\left( \cfrac { \cos { \theta  } +\sin { \theta  } +1 }{ \cos { \theta  }  }  \right) \left( \cfrac { \sin { \theta  } +\cos { \theta  } -1 }{ \sin { \theta  }  }  \right) $$
$$=\cfrac { { \left( \cos { \theta  } +\sin { \theta  }  \right)  }^{ 2 }-{ \left( 1 \right)  }^{ 2 } }{ \cos { \theta  } \sin { \theta  }  } $$
$$=\cfrac { \cos ^{ 2 }{ \theta  } +\sin ^{ 2 }{ \theta  } +2\sin { \theta  } \cos { \theta  } -1 }{ \cos { \theta  } \sin { \theta  }  } $$
$$=\cfrac { 1+2\sin { \theta  } \cos { \theta  } -1 }{ \cos { \theta  } \sin { \theta  }  } \quad =\cfrac { 2\sin { \theta \cos { \theta  }  }  }{ \cos { \theta  } \sin { \theta  }  } =2$$

we know that, $$\sec^2\theta-\tan^2\theta=1$$
$$\Rightarrow \tan^2\theta=\sec^2\theta-1$$
$$\Rightarrow \tan^2\theta=4-1$$     $$[given\,\sec\theta=2]$$
$$\Rightarrow \tan\theta=\sqrt3$$
Now,$$\displaystyle \dfrac{1-\tan \theta }{1+\tan \theta }=\dfrac{1-\sqrt3}{1+\sqrt3}\times\dfrac{1-\sqrt3}{1-\sqrt3}$$
                           $$\displaystyle=\frac{1+3-2\sqrt3}{1-3}$$
                           $$\displaystyle=\frac{4-2\sqrt3}{-2}$$
                           $$\displaystyle={\sqrt3-2}$$