CBSE Questions for Class 11 Engineering Maths Trigonometric Functions Quiz 9 - MCQExams.com

Consider the given equation.

$${{\sin }^{50}}x-{{\cos }^{50}}x=1$$

$${{\sin }^{50}}x=1+{{\cos }^{50}}x$$                 …….. (1)

Here, $${{\cos }^{50}}x$$ is always positive. So, the maximum value of $${{\sin }^{50}}x$$ is $$1$$.

So,

$${{\cos }^{50}}x$$ must be zero.

Hence, $${{\cos }^{50}}x=0$$

$$x=\left( 2n+1 \right)\dfrac{\pi }{2},$$ where $$n\in I$$

$$x=n\pi +\dfrac{\pi}{2}$$

Hence, this is the answer.

Consider the given equation,

$$\tan \left( \dfrac{\pi }{4}+\theta  \right)=3\tan 3\theta $$

Using identity ,

  $$ \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\,\,\,\And \tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A} $$

 $$ \Rightarrow \dfrac{\tan \left( \dfrac{\pi }{4} \right)+\tan \theta }{1-\tan \left( \dfrac{\pi }{4} \right)\tan \theta }=3\left( \dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right) $$

 $$ \Rightarrow \dfrac{1+\tan \theta }{1-\tan \theta }=\dfrac{9\tan \theta-3{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } $$

 $$ \Rightarrow \left( 1+\tan \theta  \right)\left( 1-3{{\tan }^{2}}\theta  \right)=\left( 1-\tan \theta  \right)\left( 9\tan \theta -3{{\tan }^{3}}\theta  \right) $$

 $$ \Rightarrow 3{{\tan }^{4}}\theta -6{{\tan }^{2}}\theta +8\tan \theta -1=0 $$

As $$a,b,c,t,=$$ are roots of this equation ,

Hence, sum of roots ,

$$\tan \left( a \right)+\tan \left( b\right)+\tan \left( c\right)+\tan \left(t\right)=$$ $$-\dfrac{coeficient\,of\,{{\tan }^{3}}\theta }{coeficient\,of\,{{\tan }^{4}}\theta }$$ =$$-\dfrac{0}{1}$$

$$\tan \left( a\right)+\tan \left(b\right)+\tan \left( c\right)+\tan \left( t \right)=0$$

Hence, this is the answer.

We have,

$$\dfrac{x}{y}=\dfrac{\cos A}{\cos B}$$

$$x=\dfrac{y\cos A}{\cos B}$$

Since,

$$=\dfrac{x\tan A+y\tan B}{x+y}$$…….. (1)

On putting the value of $$x$$ in equation (1), we get

$$ =\dfrac{\dfrac{y\cos A}{\cos B}\times \dfrac{\sin A}{\cos A}+\dfrac{y\sin B}{\cos B}}{\dfrac{y\cos A}{\cos B}+y} $$

$$ =\dfrac{\dfrac{ysinA}{\cos B}+\dfrac{y\sin B}{\cos B}}{\dfrac{y\cos A+y\cos B}{\cos B}} $$

$$ =\dfrac{\dfrac{y\left( sinA+\sin B \right)}{\cos B}}{\dfrac{y\left( \cos A+\cos B \right)}{\cos B}} $$

$$ =\dfrac{\left( sinA+\sin B \right)}{\left( \cos A+\cos B \right)} $$

We know that

$$ \sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right) $$

$$ \cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right) $$

Therefore,

$$ =\dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)}{2\cos \left( \dfrac{A+B}{2} \right)\cdot \cos \left( \dfrac{A-B}{2} \right)} $$

$$ =\tan \left( \dfrac{A+B}{2} \right) $$

Hence, the value is $$\tan \left( \dfrac{A+B}{2} \right)$$.

$$ \sin {{27}^{0}}-\cos {{27}^{0}} $$

$$ \Rightarrow \sin {{27}^{0}}-\cos \left( {{90}^{0}}-{{63}^{0}} \right) $$

$$ \Rightarrow \sin {{27}^{0}}-\sin {{63}^{0}} $$

$$ \Rightarrow 2\cos \left( \dfrac{{{27}^{0}}+{{63}^{0}}}{2} \right)\sin \left( \dfrac{{{27}^{0}}-{{63}^{0}}}{2} \right) $$

$$ \Rightarrow 2\cos {{45}^{0}}\sin \left( -{{36}^{0}} \right) $$

$$ \Rightarrow 2\times \dfrac{1}{\sqrt{2}}\times -\left[ \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right] $$

$$ \Rightarrow -\dfrac{1}{\sqrt{2}}\times \sqrt{2}\left( \dfrac{\sqrt{5-\sqrt{5}}}{2} \right) $$

$$ \Rightarrow -\dfrac{\sqrt{5-\sqrt{5}}}{2} $$

Hence, this is the answer

option $$(B)$$ is correct.

Consider the given equation.

$$ cos^2{{73}^{0}}+co{{s}^{2}}{{47}^{0}}-si{{n}^{2}}{{43}^{0}}+si{{n}^{2}}{{107}^{0}} $$

 $$ \Rightarrow co{{s}^{2}}{{73}^{0}}+co{{s}^{2}}{{47}^{0}}-si{{n}^{2}}\left( {{90}^{0}}-{{47}^{0}} \right)+si{{n}^{2}}\left( {{180}^{0}}-{{73}^{0}} \right) $$

 $$ \Rightarrow \left( co{{s}^{2}}{{73}^{0}}+{{\sin }^{2}}{{73}^{0}} \right)+co{{s}^{2}}{{47}^{0}}-{{\cos }^{2}}{{47}^{0}} $$

 $$ \Rightarrow 1 $$