MATHS M.C.Q (1 Marks)

3. $\tan\frac{\alpha}{2}$

Solution:

$\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}=\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+\cos3\alpha+2\cos4\alpha}$

$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha\cos\alpha+2\cos4\alpha}$

$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha(\cos\alpha+1)}$

$=\frac{\sin\alpha}{\cos\alpha+1}$

$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}}$

$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}}$

$=\frac{\sin\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}$

$=\tan\frac{\alpha}{2}$

1. $3\tan3\text{x}$

Solution:

$\frac{\pi}{3}=60^\circ,\frac{2\pi}{3}=120^\circ$

$\tan​​\text{x}+\tan(60^\circ+\text{x})+\tan(120^\circ+\text{x})\\=\tan\text{x}+\frac{\tan60^\circ+\tan\text{x}}{1-\tan60^\circ\tan\text{x}}+\frac{\tan120^\circ+\tan\text{x}}{1-\tan120^\circ\tan\text{x}}$

$\tan\text{x}+\frac{\sqrt{3}+\tan\text{x}}{1-\sqrt{3}\tan\text{x}}+\frac{(1\sqrt{3}+\tan\text{x})}{1+\sqrt{3}\tan\text{x}}$

$=\frac{\tan\text{x}(1-3\tan^2\text{x})+(\sqrt{3+\tan\text{x}})(1+\sqrt{3}\tan\text{x})+(-\sqrt{3}+\tan\text{x})(1-\sqrt{3}\tan\text{x})}{1-3\tan^2\text{x}}$

$=\frac{\tan\text{x}-3\tan^3\text{x}+\sqrt{3}+3\tan\text{x}+\sqrt{3}\tan^2+\tan^2\text{x}+\tan\text{x}-\sqrt{3}\tan^2\text{x}-\sqrt{3}+3\tan\text{x}}{1-3\tan^2\text{x}}$

$=\frac{9\tan\text{x}-3\tan^3\text{x}}{1-3\tan^2\text{x}}$

$=3\tan3\text{x}$

2. $\text{b}$

Solution:

Givan: $\tan\text{x}=\frac{​​\text{a}}{\text{b}}$

Now,

$=\cos2\text{x}+\alpha\sin2\text{x}$

$=\text{b}\Big(\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}\Big)+\text{a}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)$

$=\text{b}\Bigg(\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)+\text{a}\Bigg(\frac{1\times\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)$

$=\frac{\text{b}(\text{b}^2-\text{a}^2)}{\text{a}^2+\text{b}^2}+\frac{2\text{a}^2\text{b}^2}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}^3-\text{a}^2\text{b}+2\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}^3+\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}(\text{b}^2+\text{a}^2)}{\text{a}^2+\text{b}^2}$

$=\text{b}$

2. 2

Solution:

We have,

$\sin^2\Big(\frac{\pi}{18}\Big)+\sin^2\Big(\frac{7\pi}{18}\Big)+\sin^2\Big(\frac{4\pi}{9}\Big)$

$=\frac{1}{2}\Big[1-\cos\big(\frac{\pi}{9}\big)+1-\cos\big(\frac{2\pi}{9}\big)+1-\cos\frac{7\pi}{9}+1-\cos\frac{8\pi}{9}\Big]$ $\Big(\therefore\sin^2\theta=\frac{1-\cos2\theta}{2}\Big)$

$=\frac{1}{2}\Big[4-\cos\big(\frac{\pi}{9}\big)-\cos\big(\frac{2\pi}{9}\big)-\Big\{-\cos\Big(\pi-\frac{7\pi}{9}\Big)\Big\}-\big\{-\cos(\pi-\frac{8\pi}{9})\big\}\Big]$

$=\frac{1}{2}\Big[4-\cos\big(\frac{\pi}{9}\big)-\cos\big(\frac{2\pi}{9}\big)+\cos\big(\frac{2\pi}{9}\big)+\cos\big(\frac{\pi}{9}\big)\Big]$

$=\frac{4}{2}$

$=2$

1. $1$

Solution:

We have,
$\Rightarrow2\cos^2​​\text{x}-1+2\cos\text{x}=1$

$\Rightarrow\cos^2\text{x}+\cos\text{x}-1=0$

$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{1^2+4}}{2}$

$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{5}}{2}$

$\Rightarrow\cos\text{x}=\frac{-1+\sqrt{5}}{2}$

Now,

$(2-\cos^2​​\text{x})\sin^2\text{x}=\bigg[2-\Big(\frac{-1+\sqrt{5}}{2}\Big)^2\bigg](1-\cos^2\text{x})$

$=\bigg[2-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\bigg]\Big(1-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\Big)$

$=\frac{1}{4}\Big(1+\sqrt{5}\Big)\Big(\sqrt{5}-1\Big)=\frac{4}{4}=1$

2. $\sin4\beta$

Solution:

it is given thet $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}.$

Now,

$\tan\beta=\frac{2\tan\beta}{1-\tan^2\beta}$

$=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}$

$=\frac{\frac{2}{3}}{\frac{8}{9}}$

$=\frac{3}{4}$

$\therefore\tan(\alpha+2\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times\frac{3}{4}}$

$=\frac{\frac{25}{28}}{\frac{25}{28}}$

$=1$

$\tan(\alpha+2\beta)=1=\tan\frac{\pi}{4}$

$\Rightarrow\alpha+2\beta=\frac{\pi}{4}$

$\Rightarrow\alpha=\frac{\pi}{4}-2\beta$

$\Rightarrow2\alpha=\frac{\pi}{2}-4\beta$

$\Rightarrow\cos2\alpha=\cos\big(\frac{\pi}{2}-4\beta\big)=\sin4\beta$

$\therefore\cos2\alpha=\sin4\beta$

Hence, the correct answer is option B.

4.  $\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$

Solution:

Given:

$\tan\frac{​​\text{x}}{2}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}\tan\frac{\alpha}{2}$

$\Rightarrow\frac{\tan\frac{\text{x}}{2}}{\tan\frac{\alpha}{2}}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}$

Squaring both sides, we get,

$\frac{\tan^2\frac{\text{x}}{2}}{\tan^2\frac{\alpha}{2}}=\frac{1-\text{e}}{1+\text{e}}$

$\Rightarrow\tan^2\frac{\alpha}{2}(1-\text{e})=\tan^2\frac{\text{x}}{2}(1+\text{e})$

$\Rightarrow\frac{\sin^2\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}(1-\text{e})=\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}(1+\text{e})$

$\Rightarrow\frac{\frac{1}{2}(1-\cos\alpha)}{\frac{1}{2}(1+\cos\alpha)}(1-\text{e})=\frac{\frac{1}{2}(1-\cos\text{x})}{\frac{1}{2}(1+\cos\text{x})}(1+\text{e})$

$\Rightarrow(1-\cos\alpha)(1+\cos\text{x})(1-\text{e})=(1+\cos\alpha)(1-\cos\text{x})(1+\text{x})$

$\Rightarrow(1+\cos\text{x})(1-\text{e})-\cos\alpha(1+\cos\text{x})(1-\text{e})\\=(1-\cos\text{x})(1+\text{e})+\cos\alpha(1-\cos\text{x})(1+\text{e})$

$\Rightarrow\cos\alpha\big\{(1+\cos\text{x})(1-\text{e})+(1-\cos\text{x})(1+\text{e})\big\}\\=(1+\cos\text{x})(1-\text{e})-(1-\cos\text{x})(1+\text{e})$

$\Rightarrow\cos\alpha=\frac{2\cos\text{x}-2\text{e}}{2-2\text{e}\cos\text{x}}=\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$ 

3. $\cos2\text{A}$

Solution:

We have,

$2\sin^2​\text{B}+4\cos(\text{A+B})\sin​​\text{A}\sin​\text{B}+\cos2​\text{A+B}$

$=1-\cos2​\text{B}+\cos^2(​\text{A+B})+4\cos(​\text{A+B})\sin​\text{A}\sin​\text{B}$

$=1+(\cos2(​\text{A+B})-\cos2​\text{B})+4\cos(​\text{A+B})\sin​\text{A}\sin​\text{B}$

$=1-2\sin​\text{A}\sin(​\text{A+2B})+4\cos(​\text{A+B})\sin​\text{A}\sin​\text{B}\\ \ \ \ \ \ \ \ \ \Big[\because\cos​\text{C}\cos​\text{D}=-2\sin\frac{​\text{C+D}}{2}\sin\frac{​\text{C}-​\text{D}}{2}\Big]$

$=1-2\sin​\text{A}[\sin(​\text{A}+2​\text{B})-2\sin​\text{B}-2\cos(​\text{A+B})]$

$=1-2\sin​\text{A}[\sin(​\text{A}+2​\text{B})-\big\{\sin(\text{B+A+B})+\sin(\text{B}-\text{(A+B)})\big\}\\ \ \ \ \ \ [\because2\sin\text{C}\cos\text{D}=\sin(\text{C+D})+\sin(\text{C}-\text{D})]$

$=1-2\sin\text{A}\big[\sin(\text{A+2B})-\big\{\sin(\text{A+2B})+\sin(-\text{A})\big\}\big]$

$=1-2\sin\text{A}[\sin\text{A}]$

$=1-2\sin^2\text{A}$

$=\cos^2\text{A}$

3. $\cos4\text{x}$

Solution:

$\cos^4​​\text{x}+\sin^4\text{x}-6\cos^2\text{x}\sin^2\text{x}=\cos^4\text{x}\\+\sin^4\text{x}-2\cos^2\text{x}\sin^2\text{x}-4\cos^2\text{x}\sin^2\text{x}$

$=(\cos^2\text{x}-\sin^2\text{x})^2-(2\sin\text{x}\cos\text{x})^2$

$=\cos^22\text{x}-\sin^22\text{x}$

$=\cos4\text{x}$

4. 2

Solution:

Given:

$\tan\big(\frac{\pi}{4}+\text{x}+\tan\big(\frac{\pi}{4}-\text{x}\big)=\lambda\sec2\text{x}$

$\Rightarrow\frac{\tan\frac{\pi}{4}+\tan\text{x}}{1-\tan\frac{\pi}{4}\times\tan\text{x}}+\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\times\tan\text{x}}=\lambda2\text{x}$

$\Rightarrow\frac{1+\tan\text{x}}{1-\tan\text{x}}+\frac{1-\tan\text{x}}{1+\tan\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{(1+\tan\text{x})^2+(1-\tan\text{x})^2}{(1-\tan\text{x})(1+\tan\text{x})}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2(1+\tan^2\text{x})}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2\sec^2\text{x}}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos^2(1-\tan^2\text{x})}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos^2\text{x}\Big(1-\frac{\sin^2\text{x}}{\cos^2\text{x}}\Big)}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos^2\text{x}-\sin^2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow2\sec2\text{x}=\lambda\sec2\text{x}$

$\Rightarrow2=\lambda$

$\therefore\lambda=2$

3. $4\tan\beta$

Solution:

We have,

$5\sin\alpha=3\sin(\alpha+2\beta)$

$\Rightarrow\frac{5}{3}=\frac{\sin(\alpha+2\beta)}{\sin\alpha}$

$\Rightarrow\frac{5-3}{5+3}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$ (using componendo and dividendo)

$\Rightarrow\frac{2}{8}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$

$\Rightarrow\frac{1}{4}=\frac{2\cos\frac{\alpha+2\beta+\alpha}{2}\sin\frac{\alpha+2\beta-\alpha}{2}}{2\sin\frac{\alpha+2\beta+\alpha}{2}\cos\frac{\alpha+2\beta-\alpha}{2}}$

$\Rightarrow\frac{1}{4}\frac{\cos(\alpha+\beta)\sin\beta}{\sin(\alpha+\beta)\cos\beta}$

$\Rightarrow\frac{1}{4}=\cot(\alpha+\beta)\tan\beta$

$\Rightarrow\frac{1}{4}=\frac{1}{\tan(\alpha+\beta)}\tan\beta$

$\therefore\tan(\alpha+\beta)=4\tan\beta$

4. 4

Solution:

We have,

$\big(\cot\frac{\text{x}}{2}-\tan\frac{\text{x}}{2}\big)(1-2\tan\text{x}\cot2\text{x})$

$\big(\cot^2\frac{\text{x}}{2}-2\cot\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}\big)\Big\{1-2\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{2\cot\text{x}}\Big)\Big\}$

$\big(\cot^2\frac{\text{x}}{2}-2+\tan^2\frac{\text{x}}{2}\big)\Big\{1-\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{\cot\text{x}}\Big)\Big\}$

$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\Big(1-\frac{\cot^2\text{x}-\tan\text{x}}{\cot\text{x}}\Big)$

$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\big(\tan^2\text{x}\big)$

$\big(\cot^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}-2\big)\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1-\tan^2\frac{\text{x}}{2}}\Bigg)^2$

$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4+4\tan^4\frac{\text{x}}{2}-8\tan^2\frac{\text{x}}{2}\big)$

$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4-8\tan^2\frac{\text{x}}{2}+4\tan^4\frac{\text{x}}{2}\big)$

$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\Big\{\big(\tan^2\frac{\text{x}}{2}\big)^2-2\big(\tan^2\frac{\text{x}}{2}+1\big)\Big\}$

$=\frac{4\big(\tan^2\frac{​​\text{x}}{2}-1\big)^2}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}$

$=4$

4. None of these

Solution:

ABC is a tringle.

$\therefore\text{A}+\text{B}+\text{C}=\pi$

$\Rightarrow\text{A}+\text{B}+\pi-\text{C}$

$\Rightarrow\tan(\text{A}+\text{B})=\tan(\pi-\text{C})$

$\Rightarrow\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=-\tan\text{C}$

$\Rightarrow\tan\text{A}+\tan\text{B}=-\tan\text{C}+\tan\text{A}\tan\text{B}\tan\text{C}$

$\Rightarrow\tan\text{A}+\tan\text{B}+\tan\text{C}=\tan\text{A}\tan\text{B}\tan\text{C}$

$\Rightarrow0=\tan\text{A}+\tan\text{B}\tan\text{C}$ $\big[$ given: $\tan\text{A}\tan\text{B}\tan\text{C}=0\big]$

$\Rightarrow\tan\text{A}\tan\text{B}\tan\text{C}=0$

$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac{1}{0}$

$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}\rightarrow\infty$

2. $\tan2\alpha=\tan\beta$

Solution:

$\tan\alpha=\frac{1-\cos\beta}{\sin\beta}$

$=\frac{2\sin^2\frac{\beta}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}$

$=\frac{\sin\frac{\beta}{2}}{\cos\frac{\beta}{2}}$

$\Rightarrow\tan\alpha=\tan\frac{\beta}{2}$

$\Rightarrow\alpha=\frac{\beta}{2}$

$\Rightarrow2\alpha=\beta$

$\therefore\tan2\alpha=\tan\beta$

1. $16\cos^4-12\cos^2\text{x}+1$

Solution:

To find: $\frac{\sin5\text{x}}{\sin\text{x}}$

Now,

$\sin5\text{x}=\sin(3\text{x}+2\text{x})$

$=(3\sin\text{x}-4\sin^3\text{x})(1-2\sin^2\text{x})+(4\cos^3\text{x}-3\cos\text{x})(2\sin\text{x}\cos\text{x})$

$=(3\sin\text{x}-4\sin^3\text{x}-4\sin^3\text{x}+8\sin^5\text{x})+2\sin\text{x}\cos^2\text{x}(4\cos^2\text{x}-3)$

$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+2\sin\text{x}(1-\sin^2\text{x})[2(1-\sin^2\text{x})-3]$

$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-2\sin^3​​\text{x})(4-4\sin^2\text{x}-3)]$

$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-8\sin^3​​\text{x}2\sin^3\text{x}+8\sin^5​​\text{x})]$

$=5\sin\text{x}-20\sin^3+16\sin^5\text{x}$

$\therefore\frac{\sin5\text{x}}{\sin\text{x}}=\frac{5\sin\text{x}-20\sin^3\text{x}+16^5\text{x}}{\sin\text{x}}$

$=5-20\sin^2\text{x}+16\sin^4\text{x}$

$=5-20(1-\cos^2\text{x})+16(1-\cos^2\text{x})^2$

$=5-20+20\cos^2\text{x}+16(1+\cos^4\text{x}-2\cos^4\text{x})$

$=5-20+20\cos^2\text{x}+16+16\cos^4\text{x}-32\cos^2\text{x}$

$=16\cos^4-12\cos^2\text{x}+1$

1. $\sin17\text{x}-\sin11\text{x}$

Solution:

We have,

$2(1-2\sin^27​​\text{x})\sin3\text{x}=2(\cos14\text{x})\sin3\text{x}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$

$=2\sin3\text{x}\cos14\text{x}$

$=\sin17\text{x}-\sin11\text{x}$ $[\because2\sin\text{A}\cos\text{xB}=\sin(\text{A+B})-\sin(\text{A}-\text{B})]$

$\therefore2(1-2\sin^27\text{x})\sin3\text{x}=\sin17\text{x}-\sin11\text{x}$

4. $\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$

Solution:

$\therefore\cos\alpha\cos2\alpha\cos4\alpha\ ...\cos2^{\text{n}-1}\alpha=\frac{\sin2^​\text{n}\alpha​}{2^\text{n}\sin\alpha}$

3. $\text{cosec}\ \text{x}$

Solution:

We have,

$\frac{2(\sin2​\text{x}+2\cos^2\text{x}-1​)}{\cos\text{x}-\sin\text{x}-\cos3\text{x}+\sin3\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{\cos\text{x}-\sin\text{x}-4\cos^3\text{x}+3\cos\text{x}\sin\text{x}-4\sin^3\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}-4\cos^3\text{x}+2\sin\text{x}-4\sin^3\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}(1-\cos^2​​​​\text{x})+2\sin\text{x}(1-2\sin^2\text{x})}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}\sin^2\text{x}+2\sin\text{x}\cos2\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\times2\sin\text{x}\cos\text{x}\sin\text{x}+2\sin\text{x}\cos2\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin2\text{x}\sin2\text{x}+2\sin\text{x}\cos2\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin\text{x}(\sin2\text{x}+\cos2\text{x})}$

$=\frac{1}{\sin\text{x}}$

$=\text{cosec}\ \text{x}$

1. $\frac{\sqrt{5}+1}{8}$

Solution:

$\cos^248^\circ-\sin^212^\circ$

$=\cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ)$ $[\cos(\text{A+B})\cos(\text{A}-\text{B})=\cos^2\text{A}-\sin^2\text{B}]$

$=\cos60^\circ\cos36^\circ$

$=\frac{1}{2}\times\Big(\frac{\sqrt{5}+1}{4}\Big)$

$=\frac{\sqrt{5}+1}{8}$

Hence, the correct answer is option A.

1. $\frac{1+\text{t}}{1-\text{t}}$

Solution:

$\tan2​\text{x}+\sec2\text{x}=\frac{2\tan\text{x}}{1-\tan^2\text{x}}+\frac{1+\tan^2\text{x}}{1-\tan^2\text{x}}$

$=\frac{2\tan\text{x}+1\tan^2\text{x}}{1-\tan^2\text{x}}$

$=\frac{(1+\tan\text{x})^2}{1-\tan^2\text{x}}$

$=\frac{(1+\tan\text{x})(1+\tan\text{x})}{(1+\tan\text{x})(1-\tan\text{x})}$

$=\frac{1+\tan\text{x}}{1-\tan\text{x}}$

$\frac{1+\text{t}}{1-\text{t}}$ $[\tan\text{x}=\text{t}]$ (given)

2. $\tan\text{x}$

Solution:

We have,

$\cot​​\text{x}-2\cot2\text{x}=\cot\text{x}-2\frac{\cot^2\text{x}-1}{2\cot\text{x}}$

$=\frac{1}{\cot\text{x}}$

$=\tan\text{x}$

4. None of these

Solution:

We have,

$\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$

$=\frac{2\sin\frac{\pi}{65}}{2\sin\frac{\pi}{65}}\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$

$\big($dividing and multiplying by $2\sin\frac{\pi}{65}\big)$

$=\frac{2\sin\frac{2\pi}{65}}{2\times2\sin\frac{\pi}{65}}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$

$=\frac{2\sin\frac{4\pi}{65}}{2\times4\sin\frac{\pi}{65}}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$

$=\frac{2\sin\frac{8\pi}{65}}{2\times8\sin\frac{\pi}{65}}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$

$=\frac{2\sin\frac{16\pi}{65}}{2\times16\sin\frac{\pi}{65}}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}$

$=\frac{2\sin\frac{32\pi}{65}}{2\times32\sin\frac{\pi}{65}}\cos\frac{32\pi}{65}$

$=\frac{\sin\frac{64\pi}{65}}{64\sin\frac{\pi}{65}}$

$=\frac{\sin\big(\pi-\frac{\pi}{65}\big)}{64\sin\frac{\pi}{65}}$

$=\frac{\sin\frac{\pi}{65}}{64\sin\frac{\pi}{65}}$

$=\frac{1}{64}$

2. $\sin\text{x}$

Solution:

We have,

$\frac{\sin3\text{x}}{1+2\cos2\text{x}}=\frac{3\sin\text{x}-4\sin^2\text{x}}{1+2(1-2\sin^2\text{x})}$

$=\frac{3\sin\text{x}-4\sin^3\text{x}}{1+2-4\sin\text{x}}$

$=\frac{\sin\text{x}(3-4\sin^2\text{x})}{(3-4\sin^2\text{x})}$

$=\sin\text{x}$

2. $\frac{\tan8\text{A}}{\tan2\text{A}}$

Solution:

We have,

$\frac{\sec8​​\text{A}-1}{\sec4\text{A}-1}=\frac{\frac{1}{\cos2\text{A}}-1}{\frac{1}{\cos}4\text{A}-1}$

$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{1-\cos8\text{A}}{1-\cos4\text{A}}$

$=\frac{\cos4\text{A}}{\cos8\text{A}}\times\frac{2\sin^24\text{A}}{2\sin^22\text{A}}(2\sin^2\theta-1-\cos2\theta)$

$=\frac{(2\cos4\text{A}\sin4\text{A})\sin4\text{A}}{2\times\cos8\text{A}\sin^22\text{A}}$

$=\frac{\sin8\text{A}\sin4\text{A}}{\cos8\text{A}\times\text{2}\sin2\text{A}\times\sin2\text{A}}$

$=\tan8\text{A}\times\frac{2\sin2\text{A}\times\cos2\text{A}}{2\sin2\text{A}\times\sin2\text{A}}$

$=\tan8\text{A}\times\cot2\text{A}$

$=\frac{\tan8\text{A}}{\tan2\text{A}}$

3. 0

Solution:

We have,

$2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\cos^3\text{x}\sin^2\text{x}$

$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-16\Big[\frac{\cos3\text{x}+3\cos\text{x}}{4}\times\frac{(1-\cos2\text{x})}{2}\Big]$

$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[(\cos3\text{x}+3\cos\text{x})(1-\cos2\text{x})\big]$

$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2\big[\cos3\text{x}-\cos3\text{x}\cos2\text{x}+3\cos\text{x}\cos2\text{x}\big]$

$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]\\+2\cos3\text{x}\cos2\text{x}+3[2\cos\text{x}\cos2\text{x}]$

$=2\cos\text{x}-\cos3\text{x}-\cos5\text{x}-2[\cos3\text{x}+3\cos\text{x}]+\cos5\text{x}+\cos\text{x}\\\ \ \ +3\cos3\text{x}+3\cos\text{x}[2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$

$2\cos\text{x}=\cos3\text{x}-\cos5\text{x}-2\cos3\text{x}-6\cos\text{x}\\+\cos5\text{x}+\cos\text{x}+3\cos3\text{x}+3\cos\text{x}=0$

2. $1$

Solution:

Given,

$(2^\text{n}+1)\text{x}=\pi$

$\Rightarrow2^\text{n}\text{x}+\text{x}=\pi$

$\Rightarrow2^\text{n}\text{x}=\pi-\text{x}$

$\Rightarrow\sin2^\text{n}\text{x}=\sin(\pi-\text{x})$

$\Rightarrow\sin2^\text{n}\text{x}=\sin\text{x}\ .....(1)$

$2^\text{n}\cos\text{x}\cos2\text{x}\cos2^2\text{x}\ ...\cos2^{\text{n}-1}\text{x}=2^\text{n}\times\frac{\sin2^\text{n}\text{x}}{2^\text{n}\sin\text{x}}$

$=\frac{\sin2^\text{n}\text{x}}{\sin\text{x}}$

$=\frac{\sin\text{x}}{\sin\text{x}}$ [From (1)]

$=1$

1. $[-1,3]$

Solution:

$\text{A}=2\sin^2\text{x}-\cos2\text{x}$

$=2\sin^2\text{x}-(1-2\sin^2\text{x})$

$=4\sin^2\text{x}-1$

$\therefore0\leq\sin^2\text{x}\leq1$

$\Rightarrow4\times0\leq4\times\sin^2\text{x}\leq4\times1$

$\Rightarrow0\leq4\sin^2\text{x}\leq4$

$\Rightarrow0-1\leq4\sin^2\text{x}-1\leq4-1$

$\Rightarrow-1\leq4\sin^2\text{x}-1\leq3$

$\Rightarrow-1\leq\text{A}\leq3$

$\Rightarrow\text{A}\in[-1,3]$

1. $\cos2\text{A}$

Solurion:

$\cos(36^\circ-​\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ-\text{A})\cos(54^\circ+\text{A})$

$=\cos[90^\circ-(54^\circ+\text{A})]\cos[90^\circ-(54^\circ-\text{A})]+\cos(54^\circ\\-\text{A})\cos(54^\circ+\text{A})$

$=\sin(54^\circ+\text{A})\sin(54^\circ-\text{A})+\cos(54^\circ-\text{A})\cos(54^\circ+\text{A})$ $[\cos(90^\circ-\theta)=\sin\theta]$

$=\cos(54^\circ+\text{A}-54^\circ+\text{A})$ $[\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}]$

$=\cos2\text{A}$

2. $-\frac{\text{b}}{\text{a}}$

Solution:

Given:

$\sin\alpha+\sin\beta=\text{a}$

$\Rightarrow2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}=\text{a}\ .....(1)$

Also,

$\cos\alpha+\cos\beta=\text{b}$

$\Rightarrow-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha+\beta}{2}=\text{b}\ .....(2)$

On dividing (1) by (2), we get

$\frac{-\cos\frac{\alpha+\beta}{2}}{\sin\frac{\alpha-\beta}{2}}=\frac{\text{a}}{\text{b}}$

$\Rightarrow\frac{-\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}}=\frac{\text{b}}{\text{a}}$

$\Rightarrow\tan\frac{\alpha-\beta}{2}=-\frac{\text{b}}{\text{a}}$

1. $\cos\text{x}$

Solution:

We have,

$\therefore\frac{\cos3\text{x}}{2\cos2\text{x}-1}=\frac{4\cos^3​\text{x}-3\cos​\text{x}}{1(2\cos^2​\text{x}-1)-1}$ $[\therefore\cos3​\text{x}=4\cos^3​\text{x}-3\cos​\text{x}]$

$=\frac{4\cos^3​\text{x}-3\cos​\text{x}}{4\cos^2​\text{x}-2-1}$

$=\frac{4\cos^3​\text{x}-3\cos​\text{x}}{4\cos^2​\text{x}-3}$

$=\cos​\text{x}\Big(\frac{4\cos^2​\text{x}-3}{4\cos^2​\text{x}-3}\Big)$

$=\cos​\text{x}$

1. $\tan3\text{x}$

Solution:

$\frac{\pi}{3}=60^\circ$

$\tan​​\text{x}\tan(60^\circ-\text{x})\tan(60^\circ+\text{x})\\=\tan\text{x}\times\frac{\tan60^\circ-\tan\text{x}}{1+\tan60^\circ\tan\text{x}}\times\frac{\tan60^\circ+\tan\text{x}}{1-\tan60^\circ\tan\text{x}}$

$=\tan\text{x}\times\frac{\sqrt{3}-\tan\text{x}}{1+\sqrt{3\tan\text{x}}}\times\frac{\sqrt{3}+\tan\text{x}}{1-\sqrt{3}\tan\text{x}}$

$=\frac{\tan\text{x}(3-\tan^2\text{x})}{1-3\tan^2\text{x}}$

$=\frac{3\tan\text{x}-\tan^3\text{x}}{1-3\tan^2\text{x}}$

$=\tan3\text{x}$

1. $\frac{1}{2}\cos^2\text{x}$

Solution:

We have,

$\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\sin^2\Big(\frac{\pi}{6}-\text{x}\Big)$

$=\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\cos^2\big[\frac{\pi}{2}-\big(\frac{\pi}{6}-\text{x}\big)\big]$

$=\cos^2\Big(\frac{\pi}{6}+\text{x}\Big)-\cos^2\Big(\frac{\pi}{3}-\text{x}\Big)$

$=\big[\cos\big(\frac{\pi}{6}+\text{x}\big)+\cos\big(\frac{\pi}{3}+-\text{x}\big)\big]\big[\cos\big(\frac{\pi}{6}+\text{x}\big)-\cos\big(\frac{\pi}{3}+-\text{x}\big)\big]$

$=\cos\bigg(\frac{\frac{\pi}{6}+\text{x}+\frac{\pi}{3}+\text{x}}{2}\bigg)\cos\bigg(\frac{\frac{\pi}{6}+\text{x}-\frac{\pi}{3}-\text{x}}{2}\bigg)2\sin\bigg(\frac{\frac{\pi}{6}+\text{x}+\frac{\pi}{3}+\text{x}}{2}\bigg)\sin\bigg(\frac{\frac{\pi}{6}+\text{x}+\frac{\pi}{6}+\text{x}}{2}\bigg)$

$=4\cos\big(\frac{\pi}{4}+\text{x}\big)\cos\big(-\frac{\pi}{12}\big)\sin\big(\frac{\pi}{4}+\text{x}\big)\sin\big(\frac{\pi}{12}\big)$

$=4\cos\big(\frac{\pi}{4}+\text{x}\big)\cos\big(\frac{\pi}{12}\big)\sin\big(\frac{\pi}{4}+\text{x}\big)\sin\big(\frac{\pi}{12}\big)$

$=\big[2\sin\big(\frac{\pi}{4}+\text{x}\big)\cos\big(\frac{\pi}{4}+\text{x}\big)\big]\big[2\sin\big(\frac{\pi}{12}\big)\cos\big(\frac{\pi}{12}\big)\big]$

$=\sin\big(\frac{\pi}{2}+2\text{x}\big)\sin\frac{\pi}{6}$

$=\cos2\text{x}\times\frac{1}{2}$

$=\frac{1}{2}\cos2\text{x}$

4. None of these

Solution:

We have,

$\tan\theta\sin\big(\frac{\pi}{2}+\text{x}\cos\big)\big(\frac{\pi}{2}-\text{x}\big)$

$=\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}\sin\text{x}$

$=\sin^2\text{x}$

1. $0$

Solution:

We have,

$2\tan\frac{\pi}{10}+3\sec\frac{\pi}{10}-4\cos\frac{\pi}{10}$

$=2\tan18^\circ+3\sec18^\circ-4\cos18^\circ$

$=2\frac{\sin18^\circ}{\cos18^\circ}+3\times\frac{1}{\cos18^\circ}-4\cos18^\circ$

$=2\times\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}+3\times\frac{1}{\frac{\sqrt{10+2\sqrt{5}}}{4}}=4\times\frac{\sqrt{10+2\sqrt{5}}}{4}$

$=2\times\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}+3\times\frac{4}{\sqrt{10+2\sqrt{5}}}-\sqrt{10+2\sqrt{5}}$

$=\frac{2\sqrt{5}-2+12-\Big(\sqrt{10+2\sqrt{5}}\Big)^2}{\Big(\sqrt{10+2\sqrt{5}}\Big)}$

$=\frac{2\sqrt{5}+10-10-2\sqrt{5}}{\Big(\sqrt{10+3\sqrt{5}}\Big)}$

$=0$

4. $\sin\text{x}$

Solution:

we have,

$=8\sin​​\text{x}8\cos\text{x}2\cos\text{x}4\cos\text{x}8$

$=4\times2\sin\text{x}8\cos\text{x}2\cos\text{x}2$

$=4\times\sin\text{x}\cos\text{x}2\cos\text{x}4$

$=2\times2\sin\text{x}4\cos\text{x}4\cos\text{x}2$

$=2\times\sin2\cos\text{x}2$

$=\sin\text{x}$

1. $\sqrt{2}\tan\beta$

Solution:

Given:

$\cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta}$

$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{(3\cos2\beta-1)-(3-\cos2\beta)}{(3\cos2\beta-1)+(3-\cos2\beta)}$ (Using componendo and dividendo)

$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{4\cos2\beta-4}{2\cos2\beta+2}$

$\Rightarrow​-​=\frac{1-\cos^2\alpha}{1+\cos2\alpha}=\frac{-4(1-\cos2\beta)}{2(1+\cos2\beta)}$

$\Rightarrow\frac{1-\cos2\alpha}{1+\cos2\alpha}=\frac{2(1-\cos2\beta)}{(1+\cos2\beta)}$

$\Rightarrow\frac{2\sin^2\alpha}{2\cos^2\alpha}=\frac{2(2\sin^2\beta)}{2\cos^2\beta}$

$\Rightarrow\tan^2\alpha=2\tan^2\beta$

$\therefore\tan\alpha=\sqrt{2}\tan\beta$

1. $\frac{\sin2\beta}{5-\cos2\beta}$

Solution:

Given:

$2\tan\alpha=3\tan\beta$

Now,

$\tan(\alpha+\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$=\frac{\frac{3}{2}\tan\beta-\tan\beta}{1+\Big(\frac{3}{2}\tan\beta\Big)\tan\beta}$

$=\frac{3\tan\beta-2\tan\beta}{2+3\tan^2\beta}$

$=\frac{\tan\beta}{2+3\tan^2\beta}$

$=\frac{\frac{\sin\beta}{\cos\beta}}{2+3\frac{\sin^2\beta}{\cos^2\beta}}$

$=\frac{\sin\beta\cos\beta}{2\cos\beta+3\sin\beta}$

$=\frac{\sin\beta\cos\beta}{2+\sin^2\beta}$

$=\frac{2\sin\beta\cos\beta}{4+2-\cos2\beta}$

$=\frac{\sin2\beta}{4+1-cos2\beta}$

$\therefore\tan(\alpha+\beta)=\frac{\sin2\beta}{5-\cos2\beta}$

1. $\frac{1}{2}$

Solution:

Given:

$\cos\text{x}=\frac{1}{2}(\text{a}+\frac{1}{\text{a}})$

$\cos3\text{x}=\lambda\Big(\text{a}^2+\frac{1}{a^3}\Big)$

Now,

$\cos^3\text{x}=\frac{1}{8}\Big[\text{a}^3+\frac{1}{\text{a}^2}+3\text{a}\frac{1}{\text{a}^3}+\text{a}\frac{1}{\text{a}}\Big]$

$\cos^3\text{x}=\frac{1}{8}\Big(​a​^3+\frac{1}{​a​^3}+32\cos​​​\text{x}​\Big)$ $[\because\cos\text{x}=\frac{1}{2}(\text{a}+\frac{1}{\text{a}})]$

$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{\cos^3\text{x}}{\lambda}+6\cos\text{x}\Big)$

$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{4\cos^3\text{x}-3\cos\text{x}}{\lambda}+6\cos\text{x}\Big)$

$\Rightarrow\cos^3\text{x}=\frac{4\cos^3\text{x}}{8\lambda}\frac{3\cos^3\text{x}}{8\lambda}+\frac{6\cos\text{x}}{8}$

On comparing the powers of cos3x on both sides, we get

$1=\frac{4}{8\lambda}$

$1=\frac{4}{8\lambda}$

$\Rightarrow\lambda=\frac{1}{2}$