Question 13 Marks
Very-Short and Short-Answer Questions.If $5\text{x}=\sec\theta$ and $\frac{5}{\text{x}}=\tan\theta,$ find the value of $5\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big).$
AnswerGiven, $5\text{x}=\sec\theta$ and $\frac{5}{\text{x}}=\tan\theta$
We know that,
$1+\tan^2\theta=\sec^2\theta$
$\Rightarrow1+\Big(\frac{5}{\text{x}}\Big)^2=(5\text{x})^2$
$\Rightarrow25\text{x}^2-\frac{25}{\text{x}^2}=1$
$\Rightarrow25\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=1$
$\Rightarrow5\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac15$
View full question & answer→Question 23 Marks
Prove the following identities:
$\frac{1-\tan^2\theta}{\cot^2-1}=\tan^2\theta$
Answer$\text{LHS}=\frac{1-\tan^2\theta}{\cot^2\theta-1}$
$=\frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}=\frac{{\begin{pmatrix}\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}\end{pmatrix}}{}}{\begin{pmatrix}\frac{\cos^2\theta-\sin^2\theta}{\sin^2\theta}\end{pmatrix}}$
$=\Big(\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}\Big)\times\frac{\sin^2\theta}{\big(\cos^2\theta-\sin^2\theta\big)}$
$=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 33 Marks
Very-Short and Short-Answer Qustions:
Write the value of $\big(1+\tan^2\theta\big)\big(1+\sin\theta\big)\big(1-\sin\theta\big).$
Answer$\big(1+\tan^2\theta\big)\big(1+\sin\theta\big)\big(1-\sin\theta\big)$
$=\sec^2\theta\big(1-\sin^2\theta\big)$
$=\sec^2\theta\times\cos^2\theta$
$=\frac{1}{\cos^2\theta}\times\cos^2\theta$
$=1$
View full question & answer→Question 43 Marks
Prove the following identities:
$\frac{\cos\theta}{(1-\tan\theta)}+\frac{\sin^2\theta}{(\cos\theta-\sin\theta)}=(\cos\theta+\sin\theta)$
Answer$\text{L.H.S.}=\frac{\cos\theta}{(1-\tan\theta)}+\frac{\sin^2\theta}{(\cos\theta-\sin\theta)}$
$=\frac{\cos\theta}{\Big(1-\frac{\sin\theta}{\cos\theta}\Big)}+\frac{\sin^2\theta}{(\cos\theta-\sin\theta)}$
$=\frac{\cos^2\theta}{(\cos\theta-\sin\theta)}+\frac{\sin^2\theta}{(\cos\theta-\sin\theta)}$
$=\frac{\cos^2\theta-\sin^2\theta}{\big(\cos\theta-\sin\theta\big)}$
$=\frac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)}$
$=(\cos\theta+\sin\theta)$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 53 Marks
Prove the following identities:
$1+\frac{\tan^2\theta}{(1+\sec\theta)}=\sec\theta$
Answer$\text{L.H.S.}=1+\frac{\tan^2\theta}{(1+\sec\theta)}=\frac{1+\sec\theta+\tan^2\theta}{(1+\sec\theta)}$
$=\frac{\sec^2\theta+\sec\theta}{(1+\sec\theta)}$ $\Big[\because\big(1+\tan^2\theta\big)=\sec^2\theta\Big]$
$=\frac{\sec\theta(1+\sec\theta)}{(1+\sec\theta)}=\sec\theta$
$=\text{R.H.S.}$
Hence, LHS = RHS.
View full question & answer→Question 63 Marks
Very-short and Short-Answer Questions.
Write the value of $\Big(\cot^2\theta+\frac{1}{\sin^2\theta}\Big).$
Answer$\cot^2\theta+\frac{1}{\sin^2\theta}$$=\frac{\cos^2\theta}{\sin^2\theta}-\frac{1}{\sin^2\theta}$
$=\frac{\cos^2\theta-1}{\sin^2\theta}$
$=\frac{-\sin^2\theta}{\sin^2\theta}$ $\big(\sin^2\theta+\cos^2\theta=1\big)$
$=-1$
View full question & answer→Question 73 Marks
Very-Short and Short-Answer Questions:
If $\sec\theta=\text{x},$ write the value of $\tan\theta.$
Answer$\sec\theta=\text{x}$
$\Rightarrow\frac{1}{\cos\theta}=\text{x}$
$\Rightarrow\cos\theta=\frac{1}{\text{x}}$
$\Rightarrow\cos^2\theta=\frac{1}{\text{x}^2}$
$\Rightarrow\sin^2\theta=1-\cos^2\theta=1-\frac{1}{\text{x}^2}=\frac{\text{x}^2-1}{\text{x}}$
$\Rightarrow\sin\theta=\frac{\sqrt{\text{x}^2-1}}{\text{x}}$
Now, $\tan\theta=\frac{\sin\theta}{\cos\theta}$
$=\frac{\frac{\sqrt{\text{x}^2-1}}{\text{x}}}{\frac{1}{\text{x}}}=\sqrt{\text{x}^2-1}$
View full question & answer→Question 83 Marks
Very-Short and Short-Answer Questions.If $\text{cosec }\theta=2\text{x}$ and $\cot\theta=\frac{2}{\text{x}},$ find the value of $2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big).$
AnswerGiven, $\text{cosec }\theta=2\text{x}$ and $\cot\theta=\frac{2}{\text{x}}$
We know that,
$\text{cosec}^2\theta-\cot^2\theta=1$
$\Rightarrow(2\text{x})^2+\Big(\frac{2}{\text{x}}\Big)^2=1$
$\Rightarrow4\text{x}^2-\frac{4}{\text{x}^2}=1$
$\Rightarrow4\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=1$
$\Rightarrow2\times2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=1$
$\Rightarrow2\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac12$
View full question & answer→Question 93 Marks
Very-Short and Short-Answer Qustions:
Write the value of $4\tan^2\theta-\frac{4}{\cos^2\theta}.$
Answer$4\tan^2\theta-\frac{4}{\cos^2\theta}$
$=4\frac{\sin^2\theta}{\cos^2\theta}-\frac{4}{\cos^2\theta}$
$=\frac{4\sin^2\theta-4}{\cos^2\theta}$
$=\frac{4\big(\sin^2\theta-1\big)}{\cos^2\theta}$
$=\frac{4\times\big(-\cos^2\theta\big)}{\cos^2\theta}$
$=-4$
View full question & answer→Question 103 Marks
Prove the following identities:
$\frac{\cot\theta}{(\text{cosec}\theta+1)}+\frac{(\text{cosec}\theta+1)}{\cot\theta}=2\sec\theta$
Answer$\text{LHS}=\frac{\cot\theta}{(\text{cosec}\theta+1)}+\frac{(\text{cosec}\theta+1)}{\cot\theta}$
$=\frac{\big(\frac{\cos\theta}{\sin\theta}\big)}{\big(\frac{1+\sin\theta}{\sin\theta}\big)}+\frac{\big(\frac{1}{\sin\theta}+1\big)}{\big(\frac{\cos\theta}{\sin\theta}\big)}$
$=\frac{\big(\frac{\cos\theta}{\sin\theta}\big)}{\big(\frac{1+\sin\theta}{\sin\theta}\big)}+\frac{\big(\frac{1+\sin\theta}{\sin\theta}\big)}{\big(\frac{\cos\theta}{\sin\theta}\big)}$
$=\frac{\cos\theta}{1+\sin\theta}+\frac{(1+\sin\theta)}{\cos\theta}$
$=\frac{\cos^2\theta+(1+\sin\theta)^2}{\cos\theta(1+\sin\theta)}$
$=\frac{\cos^2\theta+1+\sin^2\theta+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{1+1+2\sin\theta}{\cos\theta(1+\sin\theta)}$
$=\frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}$
$=\frac{2}{\cos\theta}=2\sec\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 113 Marks
Very-Short and Short-Answer Qustions:
If $\cos\theta=\frac{2}{3},$ write the value of $\frac{(\sec\theta-1)}{(\sec\theta+1)}.$
Answer$\cos\theta=\frac{2}{3}$
$\Rightarrow\sec\theta=\frac{1}{\cos\theta}=\frac32$
$\frac{\sec\theta-1}{\sec\theta+1}=\frac{\frac32-1}{\frac32+1}$
$=\frac{\frac{3-2}{2}}{\frac{3+2}{2}}=\frac15$
View full question & answer→Question 123 Marks
Very-Short and Short-Answer Questions.If $\text{x}=\text{a}\sin\theta$ and $\text{y}=\text{b}\cos\theta,$ write the value of $\Big(\text{b}^2\text{x}^2+\text{a}^2\text{y}^2\Big).$
Answer$\text{x}=\text{a}\sin\theta$ and $\text{y}=\text{b}\cos\theta$Now,
$=\text{b}^2(\text{a}\sin\theta)^2+\text{a}^2(\text{b}\cos\theta)^2$
$=\text{a}^2\text{b}^2\sin^2\theta+\text{a}^2\text{b}^2\cos^2\theta$
$=\text{a}^2\text{b}^2\big(\sin^2\theta+\cos^2\theta\big)$
$=\text{a}^2\text{b}^2\times1$
$=\text{a}^2\text{b}^2$
View full question & answer→Question 133 Marks
Prove the following identities:
$1+\frac{\cot^2\theta}{(1+\text{cosec }\theta)}=\text{sec}\theta$
Answer$\text{L.H.S.}=1+\frac{\cot^2\theta}{(1+\text{cosec }\theta)}$
$=1+\frac{\text{cosec}^2\theta-1}{(1+\text{cosec}\theta)}$
$=\frac{1+\text{cosec}+\text{cosec}^2\theta-1^2}{1+\text{cosec}\theta}$
$=\frac{(1+\text{cosec}\theta)+(\text{cosec}\theta+1)(\text{cosec}\theta-1)}{(1+\text{cosec}\theta)}$
$=\frac{(1+\text{cosec}\theta)\big[1+\text{cosec}\theta-1)\big]}{(1+\text{cosec}\theta)}$
$=1+\text{cosec}\theta-1$
$=\text{cosec}\theta$
$=\text{R.H.S.}$
Hence, LHS = RHS.
View full question & answer→Question 143 Marks
Very-short and Short-Answer Questions.
Write the value of $\sin\theta\cos(90^\circ-\theta)+\cos\theta\sin(90^\circ-\theta).$
Answer$\sin\theta\cos(90^\circ-\theta)+\cos\theta\sin(90^\circ-\theta)$$=\sin\theta\times\sin\theta+\cos\theta\times\cos\theta$
$=\sin^2\theta+\cos^2\theta$
$=1$
View full question & answer→Question 153 Marks
Very-Short and Short-Answer Qustions:
If $\tan\theta=\frac{1}{\sqrt{5}},$ write the value of $\frac{\big(\text{cosec}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}.$
Answer$\tan\theta=\frac{1}{\sqrt{5}}\Rightarrow\tan^2\theta=\frac15$
$\Rightarrow\sec^2\theta=1+\tan^2\theta=1$
$=1+\frac{1}{5}=\frac{5+1}{5}=\frac{6}{5}$
$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{1}{\sqrt{5}}}=\sqrt{5}$
Now, $\text{cosec}^2\theta=1+\cot^2\theta$
$=1\big(\sqrt{5}\big)^2=1+5=6$
$\therefore\ \frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}=\frac{6-\frac{6}{5}}{6+\frac{6}{5}}$
$=\frac{\frac{30-6}{5}}{\frac{30+6}{5}}=\frac{24}{36}=\frac23$
View full question & answer→Question 163 Marks
Prove the following identities:
$\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=(\text{cosec}\theta+\cot\theta)$
Answer$\text{LHS}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$
$=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}}$
$=\sqrt{\frac{(1-\cos\theta)^2}{1-\cos^2\theta}}$
$=\sqrt{\frac{(1-\cos\theta)^2}{\sin^2\theta}}$
$=\frac{1+\cos\theta}{\sin\theta}$
$=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}$
$=\text{cosec}\theta-\cot\theta$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 173 Marks
Prove the following identities:
Prove that:
$\big(\sin\theta-2\sin^3\theta\big)=\big(2\cos^3\theta-\cos\theta\big)\tan\theta.$
Answer$\text{L.H.S.}=\big(\sin\theta-2\sin^3\theta\big)$
$=\sin\theta\big(1-2\sin^2\theta\big)$
$=\sin\theta(1-2\sin^2\theta)$
$\text{R.H.S}=(2\cos^3\theta-\cos\theta)\tan\theta$
$=\cos\theta\big(2\cos^2\theta-1\big)\frac{\sin\theta}{\cos\theta}$
$=\Big[2\big(1-\sin^2\theta\big)-1\Big]\sin\theta$
$=\big(1-2\sin^2\theta\big)\sin\theta$
$\Rightarrow\text{L.H.S.}=\text{R.H.S.}$
$\therefore\big(\sin\theta-2\sin^3\theta\big)=\big(2\cos^3\theta-\cos\theta\big)\tan\theta$
View full question & answer→Question 183 Marks
Very-Short and Short-Answer Qustions:
If $\cos\theta=\frac{7}{25},$ write the value of $(\tan\theta+\cot\theta).$
Answer$\cos\theta=\frac{7}{25}\Rightarrow\cos^2\theta=\frac{49}{625}$$\Rightarrow\sin^2\theta=1-\sin^2\theta=1-\frac{49}{625}=\frac{576}{625}$
$\Rightarrow\sin\theta=\frac{24}{25}$
$\tan\theta+\cot\theta$
$=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$
$=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$=\frac{1}{\sin\theta\cos\theta}$
$=\frac{1}{\frac{24}{25}\times\frac{7}{25}}$
$=\frac{625}{168}$
View full question & answer→Question 193 Marks
Prove the following identities:
$\frac{\cos^2\theta}{(1-\tan\theta)}+\frac{\sin^2\theta}{(\sin\theta-\sin\theta)}=(1+\sin\theta\cos\theta)$
Answer$\text{L.H.S.}=\frac{\cos^2\theta}{(1-\tan\theta)}+\frac{\sin^2\theta}{(\sin\theta-\sin\theta)}$ $=\frac{\cos^2\theta}{\Big(1-\frac{\sin\theta}{\cos\theta}\Big)}+\frac{\sin^3\theta}{(\sin\theta-\cos\theta)}$ $=\frac{\cos^3\theta}{(\cos\theta-\sin\theta)}+\frac{\sin^3\theta}{(\sin\theta-\cos\theta)}$$=\frac{\cos^3\theta}{(\cos\theta-\sin\theta)}-\frac{\sin^3\theta}{\cos\theta-\sin\theta}$
$=\frac{\cos^3\theta-\sin^3\theta}{(\cos\theta-\sin\theta)}$ $=\frac{(\cos\theta-\sin\theta)\big(\cos^2\theta+\cos\theta\sin\theta+\sin^2\theta\big)}{(\cos\theta-\sin\theta)}$ $\Big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\Big]$ $=(1+\cos\theta\sin\theta)$ $=\text{R.H.S.}$ $\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 203 Marks
Very-Short and Short-Answer Qustions:
If $\sqrt{3}\sin\theta=\cos\theta$ and $\theta$ is an acute angle, find the value of $\theta.$
Answer$\sqrt{3}\sin\theta=\cos\theta$
$\Rightarrow\frac{\sin\theta}{\cos\theta}=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan\theta=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan\theta=\tan30^\circ$
$\Rightarrow\theta=30^\circ$
View full question & answer→Question 213 Marks
Prove the following identities:
$\sin\theta(1+\tan\theta)+\cos\theta(1+\cot\theta)=(\sec\theta+\text{cosec }\theta)$
Answer$\text{L.H.S.}=\sin\theta(1+\tan\theta)+\cos\theta(1+\cot\theta)$
$=\Big(\sin\theta+\frac{\sin^2\theta}{\cos\theta}\Big)+\cos\theta\Big(1+\frac{\cos\theta}{\sin\theta}\Big)$
$=\Big(\sin\theta+\frac{\sin^2\theta}{\cos\theta}\Big)+\Big(\cos\theta+\frac{\cos^2\theta}{\sin\theta}\Big)$
$=\Big(\frac{\sin\theta\cos\theta+\sin^2\theta}{\cos\theta}\Big)+\Big(\frac{\cos\theta\sin\theta+\cos^2\theta}{\sin\theta}\Big)$
$=\frac{\sin\theta\cos\theta}{\cos\theta}+\frac{\sin^2\theta}{\cos\theta}+\frac{\cos\theta\sin\theta}{\sin\theta}+\frac{\cos^2\theta}{\sin\theta}$
$=\frac{\sin\theta\cos\theta}{\cos\theta}+\frac{\cos\theta\sin\theta}{\sin\theta}+\frac{\sin^2\theta}{\cos\theta}+\frac{\cos^2\theta}{\sin\theta}$
$=\sin\theta\cos\theta\Big(\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}\Big)+\frac{\sin^3\theta+\cos^3\theta}{\cos\theta\sin\theta}$
$=\sin\theta\cos\theta\Big(\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}\Big)\\+\frac{(\sin\theta+\cos\theta)\big(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta\big)}{\cos\theta\sin\theta}$ $\Big[\because\text{a}^3+\text{b}^3=(\text{a}+\text{b})\big(\text{a}^2-\text{ab}+\text{b}^2\big)\Big]$
$=(\sin\theta+\cos\theta)\Big[\frac{\sin\theta\cos\theta}{\cos\theta\sin\theta}+\frac{(1-\sin\theta\cos\theta)}{\cos\theta\sin\theta}\Big]$
$=(\sin\theta+\cos\theta)\Big[\frac{\sin\theta\cos\theta+1-\sin\theta\cos\theta}{\cos\theta\sin\theta}\Big]$
$=(\sin\theta+\cos\theta)\Big(\frac{1}{\cos\theta\sin\theta}\Big)$
$=\frac{\sin}{\cos\theta\sin\theta}+\frac{\cos\theta}{\cos\theta\sin\theta}=\sec\theta+\text{cosec }\theta$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S}$
View full question & answer→Question 223 Marks
Very-short and Short-Answer Questions.
Write the value of $\big(1+\tan^2\theta\big)\cos^2\theta.$
Answer$\big(1-\tan^2\theta\big)\cos^2\theta$$=\sec^2\theta\times\frac{1}{\sec^2\theta}$
$=1$
View full question & answer→Question 233 Marks
Very-Short and Short-Answer Qustions:
Write the value of $3\cot^2\theta-3\text{cosec}^2\theta.$
Answer$3\cot^2\theta-3\text{cosec}^2\theta.$
$=3\big(\cot^2\theta-\text{cosec}^2\theta\big)$
$=3\times(-1)$
$=-3$
View full question & answer→Question 243 Marks
Very-Short and Short-Answer Qustions:
Write the value of $\tan10^\circ\tan20^\circ\tan70^\circ\tan80^\circ.$
Answer$\tan10^\circ\tan20^\circ\tan70^\circ\tan80^\circ.$
$=\tan10^\circ\times\tan20^\circ\times\tan(90^\circ-20^\circ)\times\tan(90^\circ-10^\circ)$
$=\tan10^\circ\times\tan20^\circ\times\cot20^\circ\times\cot10^\circ$
$=(\tan10^\circ\cot10^\circ)(\tan20^\circ\cot20^\circ)$
$=1\times1$
$=1$
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