Question 12 Marks
Very-Short and Short-Answer Questions.
Find the value of $\sin48^\circ\sec42^\circ+\cos48^\circ\text{cosec }42^\circ.$
Answer$\sin48^\circ\sec42^\circ+\cos48^\circ\text{cosec }42^\circ$
$=\sin48^\circ\sec(90^\circ-48^\circ)+\cos48^\circ\text{cosec }(90^\circ-48^\circ)$
$=\sin48^\circ\text{cosec }48^\circ+\cos48^\circ\sec48^\circ$
$=\sin48^\circ\times\frac{1}{\sin48^\circ}+\cos48^\circ\times\frac{1}{\cos48^\circ}$
$=1+1$
$=2$
View full question & answer→Question 22 Marks
Very-Short and Short-Answer Qustions:
If $\cos\theta=\frac23,$ write the value of $\big(4+4\tan^2\theta\big).$
Answer$\cos\theta=\frac23\Rightarrow\cos^2\theta=\frac{4}{9}$ $\therefore\ 4\tan^2\theta=4\big(1+\tan^2\theta\big)$$=4\Big(1+\frac{\sin^2\theta}{\cos^2\theta}\Big)$
$=4\Big(\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}\Big)$
$=4\times\frac{1}{\cos^2\theta}$
$=4\times\frac{1}{\frac49}$
$=9$
View full question & answer→Question 32 Marks
Very-Short and Short-Answer Qustions:
If $\cot\text{A}=\frac43$ and $(\text{A}+\text{B})=90^\circ,$ what is the value of $\sin\text{B}?$
AnswerGiven, $\cot\text{A}=\frac43$$\text{A}+\text{B}=90^\circ$
$\Rightarrow\text{A}=90^\circ-\text{B}$
$\Rightarrow\cot\text{A}=\cot(90^\circ-\text{B})$
$\Rightarrow\cot\text{A}=\tan\text{B}$
$\Rightarrow\cot\text{A}=\tan\text{B}=\frac43$
View full question & answer→Question 42 Marks
Prove the following identities:
$\cot^2\theta-\frac{1}{\sin^2\theta}=-1$
Answer$\text{L.H.S.}=\cot^2\theta-\frac{1}{\sin^2\theta}$
$=\cot^2\theta-\text{cosec}^2\theta$
$=-1$ $\big(\text{since}1+\cot^2\theta=\text{cosec}^2\theta=-1\big)$
$=\text{R.H.S.}$
View full question & answer→Question 52 Marks
Prove that $\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}}=(\text{cosec A}+\cot\text{A}).$
Answer$\sqrt{{\frac{1+\cos\text{A}}{1-\cos\text{A}}}}=(\text{cosec }\text{A}+\cot\text{A})$$\text{LHS}=\sqrt{\frac{1+\cos\text{A}}{1-\cos\text{A}}}$
Multiplying the numerator and denominator by $(1+\cos\text{A}),$ we have:
$\sqrt{\frac{(1+\cos\text{A})^2}{(1-\cos\text{A})(1+\cos\text{A})}}$
$=\sqrt{\frac{(1+\cos\text{A})^2}{1-\cos^2\text{A}}}$
$=\frac{1+\cos\text{A}}{\sqrt{\sin^2\text{A}}}$
$=\frac{1+\cos\text{A}}{\sin\text{A}}$
$=\frac{1}{\sin\text{A}}+\frac{\cos\text{A}}{\sin\text{A}}$
$=\text{cosec }\text{A}+\cot\text{A}$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 62 Marks
Prove the following identities:
$\cos^2\theta-\frac{1}{\big(1+\cot^2\theta\big)}=1$
Answer$\text{L.H.S.}=\cos^2\theta+\frac{1}{\big(1+\cot^2\theta\big)}$
$=\cos^2\theta+\frac{1}{\text{cosec}^2\theta}$
$=\cos^2\theta+\sin^2\theta$
$=1$
$=\text{R.H.S.}$
View full question & answer→Question 72 Marks
Prove the following identities:
Show that none of the following is an identity:
$\tan^2\theta+\sin\theta=\cos^2\theta$
Answer$\tan^2\theta+\sin\theta=\cos^2\theta$
$\text{L.H.S.}=\tan^2\theta+\sin\theta$
$=\frac{\sin^2\theta}{\cos^2\theta}+\sin\theta$
$=\frac{1-\cos^2\theta}{\cos\theta}+\sin\theta$
$=\sec^2\theta-1+\sin\theta$
Since $\text{L.H.S.}\neq\text{R.H.S.}$ this is not an identity.
View full question & answer→Question 82 Marks
Prove the following identites:
$\sin^2\theta+\frac{1}{\big(1+\tan^2\theta\big)}=1$
Answer$\text{L.H.S.}=\sin^2\theta+\frac{1}{1+\tan^2\theta}$
$=\sin^2\theta+\frac{1}{\sec^2\theta}$ $\begin{bmatrix}\because\big(1-\cos^2\theta\big)=\sin^2\theta\end{bmatrix}$
$=\sin^2\theta+\cos^2\theta$
$=1$
$=\text{R.H.S.}$
$\therefore\ \text{LHS}=\text{RHS}$
View full question & answer→Question 92 Marks
Prove the following identities:
$\frac{1}{(1+\sin\theta)}+\frac{1}{(1-\sin\theta)}=2\sec^2\theta$
Answer$\text{L.H.S.}=\frac{1}{(1+\sin\theta)}+\frac{1}{(1-\sin\theta)}$
$=\frac{(1-\sin\theta)+(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}$
$=\frac{2}{1-\sin^2\theta}$
$=\frac{2}{\cos^2\theta}$
$=2\sec^2\theta$
$=\text{R.H.S.}$
View full question & answer→Question 102 Marks
Very-Short and Short-Answer Questions:
If $\sec\theta+\tan\theta=\text{x},$ find the value of $\sec\theta.$
Answer$\sec\theta+\tan\theta=\text{x}\dots(\text{i})$
Now, $\sec^2\theta-\tan^2\theta=1$
$\Rightarrow(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$
$\Rightarrow\text{x}(\sec\theta-\tan\theta)=1$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{x}}\dots(\text{ii})$
Adding (i) and (ii), we get:
$\Rightarrow\sec\theta=\frac12\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$=\frac12\Big(\frac{\text{x}^2+1}{\text{x}}\Big)=\frac{\text{x}^2+1}{2\text{x}}$
View full question & answer→Question 112 Marks
Prove the following identites:
$\frac{1}{\big(1+\tan^2\theta\big)}+\frac{1}{\big(1+\cot^2\theta\big)}=1$
Answer$\text{L.H.S.}=\frac{1}{\big(1+\tan^2\theta\big)}+\frac{1}{\big(1+\cot^2\theta\big)}$
$=\frac{1}{\sec^2\theta}+\frac{1}{\text{cosec}^2\theta}$ $\begin{bmatrix}\because\big(1-\tan^2\theta\big)=\sec^2\alpha,\\\big(1+\cot^2\theta\big)=\text{cosec}^2\theta\end{bmatrix}$
$=\cos^2\theta+\sin^2\theta$
$=1$
$=\text{R.H.S.}$
$\therefore\ \text{LHS}=\text{RHS}$
View full question & answer→Question 122 Marks
If $\sqrt{3}\tan\theta=3\sin\theta,$ prove that $\big(\sin^2\theta-\cos^2\theta\big)=\frac13.$
AnswerGiven: $\sqrt3\tan\theta=3\sin\theta$$\Rightarrow\frac{\sqrt3}{\cos\theta}=3$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$
$\Rightarrow\cos\theta=\frac{\sqrt3}{3}$
$\Rightarrow\cos^2\theta=\frac{3}{9}$
$\therefore\sin^2\theta=1-\frac{3}{9}$
$\Rightarrow\sin^2\theta=\frac{6}{9}$
$\therefore\text{LHS}=\sin^2\theta-\cos^2\theta$
$=\frac{6}{9}-\frac{3}{9}\ $ $\Big[\therefore\sin^2\theta=\frac{6}{9},\cos^2\theta=\frac{3}{9}\Big]$
$=\frac{3}{9}$
$=\frac{1}3{}$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 132 Marks
Very-Short and Short-Answer Questions.
Find the value of $\frac{\sin50^\circ}{\cos40^\circ}+\frac{\text{cosec}40^\circ}{\sec50^\circ}-4\cos50^\circ\text{cosec}40^\circ.$
Answer$\frac{\sin50^\circ}{\cos40^\circ}+\frac{\text{cosec}40^\circ}{\sec50^\circ}-4\cos50^\circ\text{cosec}40^\circ$
$=\frac{\sin(90^\circ-40^\circ)}{\cos40^\circ}+\frac{\text{cosec}(90^\circ-50^\circ)}{\sec50^\circ}\\\ -4\cos50^\circ\text{cosec}(90^\circ-50^\circ)$
$=\frac{\cos40^\circ}{\cos40^\circ}+\frac{\sec50^\circ}{\sec50^\circ}-4\cos50^\circ\sec50^\circ$
$=1+1-4\cos50^\circ\times\frac{1}{\cos50^\circ}$
$=2-4\times1$
$=-2$
View full question & answer→Question 142 Marks
Very-Short and Short-Answer Questions.
If $\tan\text{A}=\frac{5}{12},$ find thew value of $(\sin\text{A}+\cos\text{A})\sec\text{A}.$
Answer Given, $\tan\text{A}=\frac{5}{12}$
$\therefore\ (\sin\text{A}+\cos\text{A})\sec\text{A}$
$=(\sin\text{A}+\cos\text{A})\times\frac{1}{\cos\text{A}}$
$=\frac{\sin\text{A}}{\cos\text{A}}+\frac{\cos\text{A}}{\cos\text{A}}$
$=\tan\text{A}+1$
$=\frac{2}{12}+1$
$=\frac{5+12}{12}$
$=\frac{17}{12}$
View full question & answer→Question 152 Marks
Prove the following identites:
$\big(1-\cos^2\theta\big)\sec^2\theta=\tan^2\theta$
Answer$\text{L.H.S.}=\big(1-\cos^2\theta\big)\sec^2\theta$
$=\sin^2\theta\times\sec^2\theta$ $\begin{bmatrix}\because\big(1-\cos^2\theta\big)=\sin^2\theta\end{bmatrix}$
$=\sin^2\theta\times\frac{1}{\cos^2\theta}=\frac{\sin^2\theta}{\cos^2\theta}$
$=\tan^2\theta$
$=\text{R.H.S.}$
$\therefore\ \text{LHS}=\text{RHS}$
View full question & answer→Question 162 Marks
Prove that $\frac{(1+\sin\theta)}{(1-\sin\theta)}=(\sec\theta+\tan\theta)^2.$
Answer$\frac{(1+\sin\theta)}{(1-\sin\theta)}=(\sec\theta+\tan\theta)^2$$\text{LHS}=\frac{(1+\sin\theta)}{(1-\sin\theta)}$
Multiplying the numerator and denominator by $(1+\sin\theta),$ we get:
$\frac{(1+\sin\theta)^2}{1-\sin^2\theta}$
$=\frac{1+2\sin\theta+\sin^2\theta}{\cos^2\theta}$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\sec^2\theta+2\times\frac{\sin\theta}{\cos\theta}\times\sec\theta+\tan^2\theta$
$=\sec^2\theta+2\times\tan\theta\times\sec\theta+\tan^2\theta$
$=(\sec\theta+\tan\theta)^2$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 172 Marks
Prove that $(\text{cosec A}-\cot\text{A})^2=\frac{(1-\cos\text{A})}{(1+\cos\text{A})}.$
Answer$(\text{cosec }\text{A}-\cot\text{A})^2=\frac{(1-\cos\text{A})}{(1+\cos\text{A})}$$\text{LHS}=(\text{cosec }\text{A}-\cot\text{A})^2$
$=\Big(\frac{1}{\sin\text{A}}-\frac{\cos\text{A}}{\sin\text{A}}\Big)^2$
$=\Big(\frac{1-\cos\text{A}}{\sin\text{A}}\Big)^2$
$=\frac{(1-\cos\text{A})^2}{\sin^2\text{A}}$
$=\frac{(1-\cos\text{A})^2}{1-\cos^2\text{A}} $ $\Big[\because\sin^2\theta+\cos^2\theta=1\Big]$
$=\frac{(1-\cos\text{A})(1-\cos\text{A})}{(1-\cos\text{A})(1+\cos\text{A})}$
$=\frac{(1-\cos\text{A})}{(1+\cos\text{A})}=\text{RHS}$
Hence proved.
View full question & answer→Question 182 Marks
Prove the following identites:
$\big(\sec^2\theta-1\big)\cot^2\theta=1$
Answer$\text{L.H.S.}=\big(\sec^2\theta-1\big)\cot^2\theta$
$=\tan^2\theta\times\cot^2\theta$ $\Big[\because\big(\sec^2\theta-1\big)=\tan^2\theta\Big]$
$=\tan^2\theta\times\frac{1}{\tan^2\theta}$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
View full question & answer→Question 192 Marks
Very-Short and Short-Answer Qustions:
Write the value of $\frac{\tan^2\theta-\sec^2\theta}{\cot^2\theta-\text{cosec}^2\theta}.$
Answer$\frac{\tan^2\theta-\sec^2\theta}{\cot^2\theta-\text{cosec}^2\theta}$
$=\frac{-1}{-1}$ $\big(1+\tan^2\theta=\sec^2\theta$ and $1+\cot^2\theta=\text{cosec}^2\theta\big)$
$=1$
View full question & answer→Question 202 Marks
Prove the following identites:
$\big(1-\cos^2\theta\big)\text{cosec}^2\theta=1$
Answer$\text{L.H.S.}=\big(1-\cos^2\theta\big)\text{cosec}^2\theta$
$=\sin^2\theta\times\text{cosec}^2\theta$ $\Big[\because\big(1-\cos^2\theta\big)=\sin^2\theta\Big]$
$=\sin^2\theta\times\frac{1}{\sin^2\theta}$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
View full question & answer→Question 212 Marks
If $\text{cosec }\theta+\cot\theta=\text{p},$ prove that $\cos\theta=\frac{\big(\text{p}^2-1\big)}{\big(\text{p}^2+1\big)}.$
Answer$\text{cosec}\theta+\cot\theta=\text{p}$$\Rightarrow\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\text{p}$
$\Rightarrow\frac{1+\cos\theta}{\sin\theta}=\text{p}$
Squaring both sides, we get:
$\Big(\frac{1+\cos\theta}{\sin\theta}\Big)^2=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{\sin^2\theta}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{1-\cos^2\theta}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)}=\text{p}^2$
$\Rightarrow\frac{(1+\cos\theta)}{(1-\cos\theta)}=\text{p}^2$
$\Rightarrow1+\cos\theta=\text{p}^2(1-\cos\theta)$
$\Rightarrow1+\cos\theta=\text{p}^2-\text{p}^2\cos\theta$
$\Rightarrow\cos\theta\big(1+\text{p}^2\big)=\text{p}^2-1$
$\Rightarrow\cos\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$
Hence proved.
View full question & answer→Question 222 Marks
If $5\cot\theta=3,$ show that the value of $\Big(\frac{5\sin\theta-3\cos\theta}{4\sin\theta+3\cos\theta}\Big)$ is $\frac{16}{29}.$
AnswerGiven: $5\cot\theta=3$$\Rightarrow\frac{5\cos\theta}{\sin\theta}=3$ $\Big[\because\cot\theta=\frac{\cos\theta}{\sin\theta}\Big]$
$\Rightarrow5\cot\theta=3\sin\theta$
Squaring both sides, we get:
$25\cos^2\theta=9\sin^2\theta$
$\Rightarrow25\cos^2\theta=9-9\cos^2\theta$ $\Big[\because\sin^2\theta+\cos^2\theta=1\Big]$
$\Rightarrow34\cos^2\theta=9$
$\Rightarrow\cos\theta=\sqrt{\frac{9}{34}}$
$\Rightarrow\cos\theta=\frac{3}{\sqrt{34}}$
Again, $\sin^2\theta=1-\cos^2\theta$
$\Rightarrow\sin^2\theta=\frac{34-9}{34}=\frac{25}{24}$
$\Rightarrow\sin\theta=\frac{5}{\sqrt{34}}$
$\therefore\ \text{LHS}=\Big(\frac{5\sin\theta-3\cos\theta}{4\sin\theta+3\cos\theta}\Big)$
$=\frac{5\times\frac{5}{\sqrt{34}}-3\times\frac{3}{\sqrt{34}}}{4\times\frac{5}{\sqrt{34}}+3\times\frac{3}{\sqrt{34}}}$ $\Big[\because\cos\theta=\frac{3}{\sqrt{34}},\sin\theta=\frac{5}{\sqrt{34}}\Big]$
$=\frac{25-9}{20+9}$
$=\frac{16}{29}$
View full question & answer→Question 232 Marks
Prove the following identities:
Show that none of the following is an identity:
$\sin^2\theta+\sin\theta=2$
Answer$\sin^2\theta+\sin\theta=2$$\text{L.H.S}=\sin^2\theta+\sin\theta$
$=1-\cos^2\theta+\sin\theta$
$=1-\big(\cos^2\theta-\sin\theta\big)$
Since $\text{LHS}\neq\text{RHS},$ this is not an identity.
View full question & answer→Question 242 Marks
Very-Short and Short-Answer Qustions:
If $\sin\theta=\frac12,$ write the value of $\big(3\cot^2\theta+3\big).$
Answer$\sin\theta=\frac12\Rightarrow\sin^2\theta=\frac14$$\therefore\ 3\cot^2\theta+3=\frac{3\cos^2\theta}{\sin^2\theta}+3$
$=\frac{3\cos^2\theta+3\sin^2\theta}{\sin^2\theta}$
$=\frac{3\big(\cos^2\theta+\sin^2\theta\big)}{\sin^2\theta}$
$=\frac{3\times1}{\frac{1}{4}}$
$=3\times4$
$=12$
View full question & answer→Question 252 Marks
Prove the following identities:
$\text{cosec }\theta\big(1+\cos\theta\big)\big(\text{cosec }\theta-\cot\theta\big)=1$
Answer$=\Big(\text{cosec}\theta+\frac{\cos\theta}{\sin\theta}\Big)\big(\text{cosec }\theta-\cot\theta\big)$
$=\big(\text{cosec }\theta+\cot\theta\big)\big(\text{cosec }\theta-\cot\theta\big)$
$=\big(\text{cosec }^2\theta-\cot^2\theta\big)=1$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 262 Marks
Very-Short and Short-Answer Questions:
If $\sin\theta=\text{x},$ write the value of $\cot\theta.$
Answer$\sin\theta=\text{x}$
$\Rightarrow\sin^2\theta=\text{x}^2$
$\Rightarrow\cos^2\theta=1-\sin^2\theta=1-\text{x}^2$
$\Rightarrow\cos\theta=\sqrt{1-\text{x}^2}$
Now, $\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{\sqrt{1-\text{x}^2}}{\text{x}}$
View full question & answer→Question 272 Marks
Short-Answer Questions:If $2\text{x}=\sec\text{A}$ and $\frac{2}{\text{x}}=\tan\text{A},$ prove that $\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)=\frac{1}{4}.$
AnswerGiven: $\text{2x}=\sec\text{A}$$\Rightarrow\text{x}=\frac{\sec\text{A}}{2}\dots(\text{i})$
and $\frac{2}{\text{x}}=\tan\text{A}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\tan\text{A}}{2}\dots(\text{ii})$
$\therefore\text{x}+\frac{1}{\text{x}}=\frac{\sec\text{A}}{2}+\frac{\tan\text{A}}{2}$ $[\because$ From (i) and (ii)$]$
Also, $\text{x}-\frac{1}{\text{x}}=\frac{\sec\text{A}}{2}-\frac{\tan\text{A}}{2}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}-\frac{1}{\text{x}}\Big)=\Big(\frac{\sec\text{A}}{2}+\frac{\tan\text{A}}{2}\Big)\Big(\frac{\sec\text{A}}{2}-\frac{\tan\text{A}}{2}\Big)$
$\Rightarrow\text{x}^2-\frac{1}{\text{x}^2}=\frac{1}{4}\big(\sec^2\text{A}-\tan^2\text{A}\big)$
$\therefore\ \text{x}^2-\frac{1}{\text{x}^2}=\frac{1}{4}\times1\ \big(\because\sec^2\text{A}-\tan^2\text{A}=1\big)$
$=\frac{1}{4}$
Hence proved.
View full question & answer→Question 282 Marks
Very-Short and Short-Answer Qustions:
If $5\tan\theta=4,$ write the value of $\frac{\big(\cos\theta-\sin\theta\big)}{\big(\cos\theta+\sin\theta\big)}.$
Answer$5\tan\theta=4$
$\Rightarrow\tan\theta=\frac{4}{5}$
$\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\frac{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}+\frac{\sin\theta}{\cos\theta}}$
$=\frac{1-\tan\theta}{1+\tan\theta}=\frac{1-\frac{4}{5}}{1+\frac{4}{5}}$
$=\frac{\frac{5-4}{5}}{\frac{5+4}{5}}=\frac19$
View full question & answer→Question 292 Marks
Very-short and Short-Answer Questions.
Write the value of $\text{cosec}^2\theta(1+\cos\theta)(1-\sin\theta).$
Answer$\text{cosec}^2\theta(1+\cos\theta)(1-\sin\theta)$$=\text{cosec}^2\theta\big(1-\cos^2\theta\big)$
$=\text{cosec}^2\theta\times\sin^2\theta$
$=\frac{1}{\sin^2\theta}\times\sin^2\theta$
$=1$
View full question & answer→Question 302 Marks
Prove that $(\sin32^\circ\cos58^\circ+\cos32^\circ\sin58^\circ)=1.$
Answer$(\sin32^\circ\cos58^\circ+\cos32^\circ\sin58^\circ)=1$$\text{LHS}=\sin32^\circ\cos58^\circ+\cos32^\circ\sin58^\circ$
$=\sin(90^\circ-58^\circ)\cos58^\circ+\cos(90^\circ-58^\circ)\sin58^\circ$
$=\cos58^\circ\times\cos58^\circ+\sin58^\circ\times\sin58^\circ$ $\begin{bmatrix}\because\sin(90^\circ-\theta)=\cos\theta,\\\cos(90^\circ-\theta)=\cos\theta\end{bmatrix}$
$=\cos^258^\circ+\sin^258^\circ$
$=1$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\text{RHS}$
View full question & answer→Question 312 Marks
Very-Short and Short-Answer Qustions:
Write the value of $\sin^2\theta\cos^2\theta\big(1+\tan^2\theta\big)\big(1+\cot^2\theta\big)$
Answer$\sin^2\theta\cos^2\theta\big(1+\tan^2\theta\big)\big(1+\cot^2\theta\big)$
$=\sin^2\theta\times\cos^2\theta\times\sec^2\theta\times\text{cosec}^2\theta$
$=\sin^2\theta\times\cos^2\theta\times\frac{1}{\cos^2\theta}\times\frac{1}{\sin^2\theta}$
$=1$
View full question & answer→Question 322 Marks
Very-Short and Short-Answer Questions.
Write the value of $\tan1^\circ\tan2^\circ\dots\tan89^\circ.$
Answer$\tan1^\circ\tan2^\circ\dots\tan89^\circ.$
$=\big[\tan1^\circ\tan89^\circ\big]\big[\tan2^\circ\tan88^\circ\big]\dots\big[\tan44^\circ\tan66^\circ\big]\tan45^\circ$
$=\big[\tan1^\circ\tan(90^\circ-1^\circ)\big]\\\ \ \big[\tan2^\circ\tan(90^\circ-2^\circ)\big]\dots\big[\tan40^\circ\tan(90^\circ-44^\circ)\big]\dots(1)$
$=\big[\tan1^\circ\cot1^\circ\big]\big[\tan2^\circ\cot2^\circ\big]\dots\big[\tan44^\circ\cot44^\circ\big]$
$=1\times1\times\dots\times1$
$=1$
View full question & answer→Question 332 Marks
Prove the following identities:
$\tan^2\theta-\frac{1}{\cos^2\theta}=-1$
Answer$\text{L.H.S.}=\tan^2\theta-\frac{1}{\cos^2\theta}$
$=\tan^2\theta-\sec^2\theta$
$=-1$ $\big(\text{since }1+\tan^2\theta=\sec^2\theta\Rightarrow\tan^2\theta-\sec^2\theta=-1\big)$
$=\text{R.H.S.}$
View full question & answer→Question 342 Marks
Very-short and Short-Answer Questions.
Write the value of $\text{cosec}^2(90^\circ-\theta)-\tan^2\theta.$
Answer$\text{cosec}^2(90^\circ-\theta)-\tan^2\theta$$=\sec^2\theta-\tan^2\theta$
$=1$
View full question & answer→Question 352 Marks
Prove the following identites:
$\big(1+\cos\theta\big)\big(1-\cos\theta\big)\big(1+\cot^2\theta\big)=1$
Answer$\big(1-\cos^2\theta\big)\big(1+\cot^2\theta\big)$
$=\sin^2\theta\times\text{cosec}^2\theta$ $\begin{bmatrix}\because\big(1-\cos^2\theta\big)=\sin^2\theta,\\\big(1+\cot^2\theta\big)=\text{cosec}^2\theta\end{bmatrix}$
$=\sin^2\theta\times\frac{1}{\sin^2\theta}$
$=1$
$=\text{R.H.S.}$
View full question & answer→Question 362 Marks
If $2\sin\theta=\sqrt{3},$ prove that $\theta=30^\circ.$
Answer$2\sin(2\theta)=\sqrt3$$\Rightarrow\sin(2\theta)=\frac{\sqrt3}{2}$
$\Rightarrow\sin(2\theta)=\sin(60)^\circ$
$\Rightarrow2\theta=60^\circ$
$\Rightarrow\theta=\frac{60^\circ}{2}$
$\Rightarrow\theta=30^\circ$
View full question & answer→Question 372 Marks
Very-Short and Short-Answer Questions.
Write the value of $\cos1^\circ\cos2^\circ\dots\cos180^\circ.$
AnswerSince $\cos90^\circ=0$
$\cos1^\circ\cos2^\circ\cos30^\circ\dots\cos90^\circ\dots\cos180^\circ=0$
View full question & answer→Question 382 Marks
Very-Short and Short-Answer Questions.
If $\sin\theta=\cos(\theta-45^\circ),$ where $\theta$ is acute, find the value of $\theta.$
Answer$\sin\theta=\cos(\theta-45^\circ)$
$\Rightarrow\cos(90^\circ-\theta)=\cos(\theta-45^\circ)$
$\Rightarrow90^\circ-\theta=\theta-45^\circ$
$\Rightarrow2\theta=135^\circ$
$\Rightarrow\theta=67.5^\circ$
View full question & answer→Question 392 Marks
Very-short and Short-Answer Questions.
Write the value of $\Big(\sin^2\theta+\frac{1}{1+\tan^2\theta}\Big).$
Answer$\sin^2\theta+\frac{1}{1+\tan^2\theta}$$=\sec^2\theta+\frac{1}{\sec^2\theta}$
$=\sin^2\theta+\frac{1}{\frac{1}{\cos^2\theta}}$
$=\sin^2\theta+\cos^2\theta$
$=1$
View full question & answer→Question 402 Marks
If $\text{x}=\text{a}\sin\theta+\text{b}\cos\theta$ and $\text{y}=\text{a}\cos\theta-\text{b}\sin\theta,$ prove that $\text{x}^2+\text{y}^2=\text{a}^2+\text{b}^2.$
AnswerGiven: $\text{x}=\text{a}\sin\theta+\text{b}\cos\theta$Squaring both sides, we get:
$\text{x}^2=\text{a}^2\sin^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\cos^2\theta\dots(\text{i})$
Also, $\text{y}=\text{a}\cos\theta-\text{b}\sin\theta$
Squaring both sides, we get:
$\text{y}^2=\text{a}^2\cos^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\dots(\text{ii})$
$\therefore\ \text{LHS}=\text{x}^2+\text{y}^2$
$=\text{a}^2\sin^2\theta+\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\\+\text{a}^2\cos^2\theta-\text{2ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta$ [Using (i) and (ii)]
$=\text{a}^2\big(\sin^2\theta+\cos^2\theta\big)+\text{b}^2\big(\sin^2\theta+\cos^2\theta\big)$
$=\text{a}^2+\text{b} ^2$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 412 Marks
Very-short and Short-Answer Questions.
Write the value of $\big(1-\cos^2\theta\big)\text{cosec}^2\theta.$
Answer$\big(1-\cos^2\theta\big)\text{cosec}^2\theta$$=\sin^2\theta\times\frac{1}{\sin^2\theta}$
$=1$
View full question & answer→Question 422 Marks
Prove the following identites:
$\big(1+\cot^2\theta\big)\sin^2\theta=1$
Answer$\text{L.H.S.}=\big(1+\cot^2\theta\big)\sin^2\theta$
$=\text{cosec}^2\times\sin^2\theta$ $\Big[\because\big(1+\cot^2\theta\big)=\text{cosec}^2\theta\Big]$
$=\frac{1}{\sin^2\theta}\times\sin^2\theta$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
View full question & answer→Question 432 Marks
Very-short and Short-Answer Questions.
Write the value of $\big(1+\cot^2\theta\big)\sin^2\theta.$
Answer$\big(1-\cot^2\theta\big)\sin^2\theta$$=\text{cosec}^2\theta\times\frac{1}{\text{cosec}^2\theta}$
$=1$
View full question & answer→Question 442 Marks
Prove the following identites:
$\big(\sec^2\theta-1\big)\big(\text{cosec}^2\theta-1\big)=1$
Answer$\text{L.H.S.}=\big(\sec^2\theta-1\big)\big(\text{cosec}^2\theta-1\big)$
$=\tan^2\theta\times\cot^2\theta$ $\begin{bmatrix}\because\big(\sec^2\theta-1\big)=\tan^2\theta,\\\big(\text{cosec}^2\theta-1\big)=\cot^2\theta\end{bmatrix}$
$=\tan^2\theta\times\frac{1}{\tan^2\theta}$
$=1$
$=\text{R.H.S.}$
$\therefore\text{LHS}=\text{RHS}$
View full question & answer→Question 452 Marks
Very-Short and Short-Answer Qustions:
If $\cot\theta=\frac{1}{\sqrt{3}},$ write the value of $\frac{\big(1-\cos^2\theta\big)}{\big(2-\sin^2\theta\big)}.$
Answer$\cot\theta=\frac{1}{\sqrt{3}}\Rightarrow\cot^2\theta=\frac13$
Now, $1+\cot^2\theta=\text{cosec}^2\theta$
$\Rightarrow\text{cosec}^2\theta=1+\frac13=\frac{3+1}{3}=\frac43$
$\Rightarrow\sin^2\theta=\frac{1}{\text{cosec}^2\theta}=\frac34$
$=\frac{1-\cos^2\theta}{2-\sin^2\theta}=\frac{\sin^2\theta}{2-\sin^2\theta}$
$=\frac{\frac{3}{4}}{2-\frac{3}{4}}=\frac{\frac34}{\frac{8-3}{4}}$
$=\frac{\frac34}{\frac54}=\frac35$
View full question & answer→Question 462 Marks
Very-Short and Short-Answer Questions:Find the value of $\frac{\cos38^\circ\text{cosec }52^\circ}{\tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan55^\circ}.$
Answer$\frac{\cos38^\circ\text{cosec }52^\circ}{\tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan55^\circ}.$
$=\frac{\cos38^\circ\text{cosec }(90^\circ-30^\circ)}{\tan18^\circ\tan35^\circ\big(\sqrt{3}\big)\tan(90^\circ-18^\circ)\tan(90^\circ-35^\circ)}$
$=\frac{\cos38^\circ\sec38^\circ}{\tan18^\circ\tan35^\circ\big(\sqrt{3}\big)\cot18^\circ\cot35^\circ}$
$=\frac{\cos38^\circ\times\frac{1}{\cos38^\circ}}{(\tan18^\circ\cot18^\circ)(\tan35^\circ\cot35^\circ)\big(\sqrt{3}\big)}$
$=\frac{1}{\sqrt{3}}$
View full question & answer→Question 472 Marks
Very-short and Short-Answer Questions.
Write the value of $\big(1-\sin^2\theta\big)\sec^2\theta.$
Answer$\big(1-\sin^2\theta\big)\sec^2\theta$$=\cos^2\theta\times\frac{1}{\cos^2\theta}$
$=1$
View full question & answer→Question 482 Marks
Very-Short and Short-Answer Qustions:
If $3\cot\theta=4,$ write the value of $\frac{\big(2\cos\theta+\sin\theta\big)}{\big(4\cos\theta-\sin\theta\big)}.$
Answer$3\cot\theta=4$
$\Rightarrow\cot\theta=\frac{4}{3}$
$\frac{2\cos\theta+\sin\theta}{4\cos\theta-\sin\theta}=\frac{\frac{2\cos\theta}{\sin\theta}+\frac{\sin\theta}{\sin\theta}}{\frac{4\cos\theta}{\sin\theta}-\frac{\sin\theta}{\sin\theta}}$
$=\frac{2\cot\theta+1}{4\cot\theta-1}=\frac{2\times\frac{4}{3}+1}{4\times\frac{4}{3}-1}$
$=\frac{\frac{8}{3}+1}{\frac{16}{3}-1}=\frac{\frac{8+3}{3}}{\frac{16-3}{3}}$
$=\frac{11}{13}$
View full question & answer→Question 492 Marks
Short-Answer Questions:If $\cot\text{A}=\frac{4}{5}$ prove that $\frac{(\sin\text{A}+\cos\text{A})}{(\sin\text{A}-\cos\text{A})}=9$
AnswerGiven: $\cot\text{A}=\frac{4}{5}$Writing $\cot\text{A}=\frac{\cos\text{A}}{\sin\text{A}}$ and squaring the equation, we get:
$\frac{\cos^2\text{A}}{\sin^2\text{A}}=\frac{16}{25}$
$\Rightarrow25\cos^2\text{A}=16\sin^2\text{A}$
$\Rightarrow25\cos^2\text{A}=16-16\cos^2\text{A}$
$\Rightarrow\cos^2\text{A}=\frac{16}{41}$
$\Rightarrow\cos\text{A}=\frac{4}{\sqrt{41}}$
$\therefore\sin^2\text{A}=1-\cos^2\text{A}$
$1-\frac{16}{41}$
Now, $\sin\text{A}=\sqrt{\frac{25}{41}}$
$\Rightarrow\sin\text{A}=\frac{5}{\sqrt{41}}$
$\therefore\text{LHS}=\frac{\sin\text{A}+\cos\text{A}}{\sin\text{A}-\cos\text{A}}$
$=\frac{\frac{5}{\sqrt{41}}+\frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}}$
$=\frac{9}{1}$
$=9=\text{RHS}$
View full question & answer→Question 502 Marks
Prove that $\frac{\big(\sin^273^\circ+\sin^217\big)}{\big(\cos^228^\circ+\cos^262^\circ\big)}=1.$
Answer$\frac{\big(\sin^273^\circ+\sin^217^\circ\big)}{\big(\cos^228^\circ+\cos^262^\circ\big)}=1$$\text{LHS}=\frac{\sin^273^\circ+\sin^217^\circ}{\cos^228^\circ+\cos^262^\circ}$
$=\frac{\big[\sin(90^\circ-17^\circ)\big]^2+\sin^217^\circ}{\big[\cos(90^\circ-62^\circ)\big]^2+\cos^262^\circ}$
$=\frac{1}{1}$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=1=\text{RHS}$
View full question & answer→Question 512 Marks
Very-short and Short-Answer Questions.
Write the value of $\sec^2\theta(1+\sin\theta)(1-\sin\theta).$
Answer$\sec^2\theta(1+\sin\theta)(1-\sin\theta)$$=\sec^2\theta\big(1-\sin^2\theta\big)$
$=\sec^2\theta\times\cos^2\theta$
$=\frac{1}{\cos^2\theta}\times\cos^2\theta$
$=1$
View full question & answer→Question 522 Marks
Very-Short and Short-Answer Qustions:
If $\cos\text{B}=\frac35$ and $(\text{A}+\text{B})=90^\circ,$ what is the value of $\sin\text{A}.$
AnswerGiven, $\cos\text{B}=\frac35$$\text{A}+\text{B}=90^\circ$
$\Rightarrow\text{B}=90^\circ-\text{A}$
$\Rightarrow\cos\text{B}=\cot(90^\circ-\text{A})$
$\Rightarrow\cos\text{B}=\sin\text{A}$
$\Rightarrow\cos\text{B}=\sin\text{A}=\frac35$
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