Exemplar Solution

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:\\ sin\frac{\pi }{10}.sin\frac{13\pi }{10}=sin\frac{\pi }{10}.sin\left(\pi +\frac{3\pi }{10}\right)\\ =sin\frac{\pi }{10}.\left(-sin\frac{3\pi }{10}\right)=-sin 18^0.sin 54^0\\ =-sin 18^0.sin \left(90^0-36^0\right)=-sin 18^0.cos36^0\\ =-\left(\frac{\sqrt[]{5}-1}{4}\right)\left(\frac{\sqrt[]{5}+1}{4}\right)\left[?sin 18^0= \frac{\sqrt[]{5}-1}{4},cos36^0=\frac{\sqrt[]{5}+1}{4}\right]\\ =-\left(\frac{5-1}{16}\right)=-\frac{1}{4}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:z=x+iy\\ Toprove:z\overline{z}+2\left(z+\overline{z}\right)+b=0\\ \Rightarrow \left(x+iy\right)\left(x-iy\right)+2\left(x+iy+x-iy\right)+b=0\\ \Rightarrow x^2+y^2-2\left(x+x\right)+b=0\\ \Rightarrow x^2+y^2-4x+b=0\\ Whichrepresentsacircle.Henceproved.\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

         $\begin{array}{l}Giventhat:tanA=\frac{1}{2}andtanB=\frac{1}{3}\\ tan2A=\frac{2tanA}{1-ta{n}^{2}A}=\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}\\ =\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\\ So,tan2A=\frac{4}{3}andtanB=\frac{1}{3}\\ tan\left(2A+B\right)=\frac{tan2A+tanB}{1-tan2A.tanB}=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3}\times \frac{1}{3}}=\frac{\frac{5}{3}}{\frac{9-4}{9}}=\frac{5}{3}\times \frac{9}{5}=3\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

        $\begin{array}{l}Giventhat:cos12^0+cos84^0+cos156^0+cos132^0\\ =\left(cos132^0+cos12^0\right)+\left(+cos156^0+cos84^0\right)\\ =\left(2cos\frac{132^0+12^0}{2}.cos\frac{132^0-12^0}{2}\right)+\left(2cos\frac{156^0+84^0}{2}.cos\frac{156^0-84^0}{2}\right)\\ =2cos72^0.cos60^0+2cos120^0.cos36^0\\ =2cos72^0\times \frac{1}{2}+2\times \left(-\frac{1}{2}\right)cos36^0=cos72^0-cos36^0\\ =cos\left(90^0-18^0\right)-cos36^0=sin18^0-cos36^0\\ =\frac{\sqrt[]{5}-1}{4}-\frac{\sqrt[]{5}+1}{4}\left[?sin18^0=\frac{\sqrt[]{5}-1}{4},cos36^0=\frac{\sqrt[]{5}+1}{4}\right]\\ =\frac{\sqrt[]{5}-1-\sqrt[]{5}-1}{4}=\frac{-1}{2}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:cos2\theta cos2?+si{n}^2\left(\theta -?\right)-si{n}^2\left(\theta +?\right)\\ cos2\theta cos2?+si{n}^2\left(\theta -?\right)-si{n}^2\left(\theta +?\right)=cos2\theta cos2?+sin\left(\theta -?+\theta +?\right).sin\left(\theta -?-\theta -?\right)\\ \left[?si{n}^{2}A-si{n}^{2}B=sin\left(A+B\right).sin\left(A-B\right)\right]\\ =cos2\theta cos2?+sin2\theta .sin\left(-2?\right)\\ =cos2\theta cos2?-sin2\theta .sin2?\left[?sin\left(-\theta \right)=-sin\theta \right]\\ =cos\left(2\theta +2?\right)=cos2\left(\theta +?\right)\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:\left(1+i\right)z=\left(1-i\right)\overline{z}\\ \Rightarrow \frac{z}{\overline{z}}=\frac{1-i}{1+i}=\frac{1-i}{1+i}\times \frac{1-i}{1-i}=\frac{1+i^2-2i}{1-i^2}\\ =\frac{1-1-2i}{1+1}=\frac{-2i}{2}=-i\\ \Rightarrow \frac{z}{\overline{z}}=-i\\ \therefore z=-i.\overline{z}Henceproved.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:a=cos\theta +isin\theta \\ \therefore \frac{1+a}{1-a}=\frac{1+cos\theta +isin\theta }{1-cos\theta -isin\theta }\\ =\frac{1+cos\theta +isin\theta }{1-cos\theta -isin\theta }\times \frac{1-cos\theta +isin\theta }{1-cos\theta +isin\theta }\\ =\frac{1-cos\theta +isin\theta +cos\theta -co{s}^2\theta +isin\theta cos\theta +isin\theta -isin\theta cos\theta +i^{2}si{n}^2\theta }{{\left(1-cos\theta \right)}^2-i^{2}si{n}^2\theta }\\ =\frac{1+isin\theta -co{s}^2\theta +isin\theta -si{n}^2\theta }{1+co{s}^2\theta -2cos\theta +si{n}^2\theta }\\ =\frac{si{n}^2\theta +2isin\theta -si{n}^2\theta }{1+1-2cos\theta }=\frac{2isin\theta }{2-2cos\theta }=\frac{2isin\theta }{2\left(1-cos\theta \right)}=\frac{isin\theta }{1-cos\theta }\\ =\frac{2sin\frac{\theta }{2}cos\frac{\theta }{2}.i}{2si{n}^2\frac{\theta }{2}}=cot\frac{\theta }{2}.i\\ Hence,\frac{1+a}{1-a}=i.cot\frac{\theta }{2}.\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenexpressioniscot\left(\frac{\pi }{4}+\theta \right).cot\left(\frac{\pi }{4}-\theta \right)\\ =\frac{cot\frac{\pi }{4}cot\theta -1}{cot\theta +cot\frac{\pi }{4}}\times \frac{cot\frac{\pi }{4}cot\theta +1}{cot\theta -cot\frac{\pi }{4}}\\ =\frac{1.cot\theta -1}{cot\theta +1}\times \frac{1.cot\theta +1}{cot\theta -1}\\ =\frac{cot\theta -1}{cot\theta +1}\times \frac{cot\theta +1}{cot\theta -1}=1\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Wehave{\left(\frac{1-i}{1+i}\right)}^{100}=a+bi\\ \Rightarrow {\left(\frac{1-i}{1+i}\times \frac{1-i}{1-i}\right)}^{100}=a+bi\Rightarrow {\left(\frac{1+i^2-2i}{1-i^2}\right)}^{100}=a+bi\\ \Rightarrow {\left(\frac{1-1-2i}{1+1}\right)}^{100}=a+bi\Rightarrow {\left(\frac{-2i}{2}\right)}^{100}=a+bi\\ \Rightarrow {\left(-i\right)}^{100}=a+bi\Rightarrow i^{100}=a+bi\\ \Rightarrow {\left(i^4\right)}^{25}=a+bi\Rightarrow {\left(1\right)}^{25}=a+bi\Rightarrow 1=a+bi\\ \Rightarrow 1+0i=a+bi\\ Comparingtherealandimaginaryparts,wegeta=1,b=0.\\ Hence,\left(a,b\right)=\left(1,0\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenexpressionissin\left(45^0+\theta \right)-cos\left(45^0-\theta \right)\\ sin\left(45^0+\theta \right)=sin45^{0}cos\theta +cos45^{0}sin\theta \\ =\frac{1}{\sqrt[]{2}}cos\theta +\frac{1}{\sqrt[]{2}}sin\theta \\ cos\left(45^0-\theta \right)=cos45^{0}cos\theta +sin45^{0}sin\theta \\ =\frac{1}{\sqrt[]{2}}cos\theta +\frac{1}{\sqrt[]{2}}sin\theta \\ sin\left(45^0+\theta \right)-cos\left(45^0-\theta \right)=\frac{1}{\sqrt[]{2}}cos\theta +\frac{1}{\sqrt[]{2}}sin\theta -\frac{1}{\sqrt[]{2}}cos\theta -\frac{1}{\sqrt[]{2}}sin\theta =0\\ Hence,thecorrectoptionis\left(d\right).\end{array}$