Question 15 Marks
Prove the following identities:
$\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}=\tan\text{A}\tan\text{B}$
Answer$\text{LHS}=\frac{\tan\text{A}+\tan\text{B}}{\cot\text{A}+\cot\text{B}}$
$=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\cos\text{A}}{\sin\text{A}}+\frac{\cos\text{B}}{\sin\text{ B}}}$
$=\frac{\frac{\sin\text{B}\cos\text{B}+\sin\text{B}\cos\text{A}}{\cos\text{A}\cos\text{B}}}{\frac{\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}}{\sin\text{A}\sin\text{B}}}$
$=\frac{(\sin\text{A}\cos\text{B}+\sin\text{B}\cos\text{A})\times\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}\times\big(\cos\text{A}\sin\text{B}+\cos\text{B}\sin\text{A}\big)}$
$=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}$
$=\tan\text{A}\tan\text{B}$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 25 Marks
If $\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)=1$ and $\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)=1,$ prove that $\Big(\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}\Big)=2.$
AnswerWe have $\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)=1$
Squaring both side, we have:
$\Big(\frac{\text{x}}{\text{a}}\sin\theta-\frac{\text{y}}{\text{b}}\cos\theta\Big)^2=(1)^2$
$\Rightarrow\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta-2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)=1\dots(\text{i})$
Again, $\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)=1$
Squaring both side, we get:
$\Big(\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta\Big)^2=(1)^2$
$\Rightarrow\Big(\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta+2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)=1\dots(\text{ii})$
Now, adding (i) and (ii), we get:
$\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta-2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)\\ \ +\Big(\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta+2\frac{\text{x}}{\text{a}}\times\frac{\text{y}}{\text{b}}\sin\theta\cos\theta\Big)=2$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{b}^2}\cos^2\theta-\frac{\text{x}^2}{\text{a}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta=2$
$\Rightarrow\Big(\frac{\text{x}^2}{\text{a}^2}\sin^2\theta+\frac{\text{y}^2}{\text{a}^2}\cos^2\theta\Big)+\Big(\frac{\text{y}^2}{\text{b}^2}\cos^2\theta+\frac{\text{y}^2}{\text{b}^2}\sin^2\theta\Big)=2$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}\big(\sin^2\theta+\cos^2\theta\big)+\frac{\text{y}^2}{\text{b}^2}\big(\cos^2+\sin^2\theta\big)=2$
$\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2$ $\big[\because\ \sin^2\theta+\cos^2\theta=1\big]$
$\therefore\ \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=2$
View full question & answer→Question 35 Marks
If $(\tan\theta+\sin\theta)=\text{m}$ and $(\tan\theta-\sin\theta)=\text{n},$ prove that $\big(\text{m}^2-\text{n}^2\big)^2=16\text{mn}.$
AnswerWe have $(\tan\theta+\sin\theta)=\text{m}$ and $(\tan\theta-\sin\theta)=\text{n}$
Now, $\text{LHS}=\big(\text{m}^2-\text{n}^2\big)^2$
$=\Big[(\tan\theta+\sin^2\theta)-(\tan\theta-\sin\theta)^2\Big]^2$
$=\Big[(\tan^2\theta+\sin^2\theta+2\tan\theta\sin\theta\\ \ \ -\big(\tan^2\theta-\sin^2\theta+2\tan\theta\sin\theta\big)\Big]^2$
$=\Big[(\tan^2\theta+\sin^2\theta+2\tan\theta\sin\theta\\ \ \ -\tan^2\theta-\sin^2\theta+2\tan\theta\sin\theta\big)\Big]^2$
$=\big(4\tan\theta\sin\theta\big)^2$
$=16\tan^2\theta\sin^2\theta$
$=16\frac{\sin^2\theta}{\cos^2\theta}\sin^2\theta$
$=16\frac{\big(1-\cos^2\theta\big)\sin^2\theta}{\cos^2\theta}$
$=16\big[\tan^2\theta\big(1-\cos^2\theta\big)\big]$
$=16\big(\tan^2\theta-\sin^2\theta\big)$
$=16(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)$
$=16\text{mn}$ $\big[(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)=\text{mn}\big]$
$\therefore\ \big(\text{m}^2-\text{n}^2\big)\big(\text{m}^2-\text{n}^2\big)=16\text{mn}$
View full question & answer→Question 45 Marks
If $\sec\theta+\tan\theta=\text{p},$ prove that:$\sec\theta=\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)$
AnswerWe have,$\sec\theta+\tan\theta=\text{p}\dots(1)$
$\Rightarrow\frac{\sec\theta+\tan\theta}{1}\times\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{1}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{p}}\dots(2)$
Adding (1) and (2), we get:
$2\sec\theta=\text{p}+\frac{1}{\text{p}}$
$\Rightarrow\sec\theta=\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)$
View full question & answer→Question 55 Marks
If $(\text{cosec }\theta+\sin\theta)=\text{a}^3$ and $(\sec\theta-\cos\theta)=\text{b}^3,$ prove that $\big(\text{a}^2\text{b}^2\big)\big(\text{a}^2+\text{b}^2\big)=1.$
AnswerWe have $(\text{cosec }\theta-\sin\theta)=\text{a}^3$ and $(\sec\theta-\cos\theta)=\text{n}$
$\Rightarrow\text{a}^3=\Big(\frac{1}{\sin\theta}-\sin\theta\Big)$
$\Rightarrow\text{a}^3=\frac{\big(1-\sin^2\theta\big)}{\sin\theta}$
$=\frac{\cos^2\theta}{\sin\theta}$
$\therefore\ \text{a}=\frac{\cos^{\frac23}\theta}{\sin^\frac13\theta}$
Again, $(\sec\theta-\cos\theta)=\text{b}^3$
$\Rightarrow\text{b}^3=\Big(\frac{1}{\cos\theta}-\cos\theta\Big)$
$=\frac{\big(1-\cos^2\theta\big)}{\cos\theta}$
$=\frac{\sin^2\theta}{\cos\theta}$
$\therefore\ \text{b}=\frac{\sin^\frac{2}{3}\theta}{\cos^\frac13\theta}$
Now, $\text{LHS}=\text{a}^2\text{b}^2\big(\text{a}^2+\text{b}^2\big)$
$=\text{a}^4\text{b}+\text{a}^2\text{b}^4$
$=\text{a}^3\big(\text{ab}^2\big)+\big(\text{a}^2\text{b}\big)\text{b}^3$
$=\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\\ \ +\frac{\cos^3\theta}{\sin\theta}\times\Bigg[\frac{\cos^\frac23\theta}{\sin^\frac13\theta}\times\frac{\sin^\frac43\theta}{\cos^\frac23\theta}\Bigg]\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\cos^2\theta}{\sin\theta}\times\sin\theta+\cos\theta\times\frac{\sin^2\theta}{\cos\theta}$
$=\cos^2\theta+\sin^2\theta=1$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 65 Marks
If $\cos\theta+\sin\theta)=\sqrt{2}\sin\theta,$ show that $\sin\theta-\cos\theta=\sqrt{2}\cos\theta.$
AnswerGiven:$\cos\theta+\sin\theta=\sqrt{2}\sin\theta$
We have $(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\big(\sin^2\theta+\cos^2\theta\big)$
$\Rightarrow\big(\sqrt{2}\sin\theta\big)^2+(\sin\theta-\cos\theta)^2=2$
$\Rightarrow2\sin^2\theta+(\sin\theta-\cos\theta)^2=2$
$\Rightarrow(\sin\theta-\cos\theta)^2=2-2\sin^2\theta$
$\Rightarrow(\sin\theta-\cos\theta)^2=2\big(1-\sin^2\theta\big)$
$\Rightarrow(\sin\theta-\cos\theta)^2=2\cos^2\theta$
$\Rightarrow(\sin\theta-\cos\theta)=\sqrt{2}\cos\theta$
Hence proved.
View full question & answer→Question 75 Marks
Prove the following identities:
$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
Answer$\text{LHS}=\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}$
On dividing numerator and denominator of LHS $\cos^\theta,$
We, get
$\text{LHS}=\frac{\sec\theta+1+\tan\theta}{\sec\theta+1-\tan\theta}$
$=\frac{(\sec\theta+\tan\theta)+\big(\sec^2\theta-\tan^2\theta\big)}{1+\sec\theta-\tan\theta}$
Writing 1 $=\big(\sec^2\theta-\tan^2\theta\big)$
$=\frac{(\sec\theta+\tan\theta)+(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta-\tan\theta)}$
$=\frac{(\sec\theta+\tan\theta)(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta-\tan\theta)}$
$=(\sec\theta+\tan\theta)=\Big(\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\Big)$
$=\Big(\frac{1+\sin\theta}{\cos\theta}\Big)$
$=\text{R.H.S.}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 85 Marks
If $\sec\theta+\tan\theta=\text{p},$ prove that:$\tan\theta=\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)$
AnswerWe have,$\sec\theta+\tan\theta=\text{p}\dots(1)$
$\Rightarrow\frac{\sec\theta+\tan\theta}{1}\times\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{1}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{p}}\dots(2)$
Subtracting (1) and (2), we get:
$2\tan\theta=\Big(\text{p}-\frac{1}{\text{p}}\Big)$
$\Rightarrow\tan\theta=\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)$
View full question & answer→Question 95 Marks
If $(\text{cosec}\theta+\cot\theta)=\text{m}$ and $(\text{cosec }\theta-\cot\theta)=\text{n},$ show that mn = 1.
AnswerWe have $(\text{cosec }\theta+\cot\theta)=\text{m}\dots(\text{i})$
Again, $(\text{cosec }\theta-\cot\theta)=\text{n}\dots(\text{ii})$
Now, multiplying (i) and (ii), we get:
$(\text{cosec }\theta+\cot\theta)\times(\text{cosec }\theta-\cot\theta)=\text{mn}$
$\Rightarrow\text{cosec}^2\theta-\cot^2\theta=\text{mn}$
$\Rightarrow1=\text{mn}$ $\big[\because\text{cosec}^2\theta-\cot^2\theta=1\big]$
$\therefore\ \text{mn}=1$
View full question & answer→Question 105 Marks
If $(2\sin\theta+3\cos\theta)=2,$ and $(3\sin\theta-2\cos\theta)=\pm3.$
AnswerGiven: $(2\sin\theta-\cos\theta)=2\dots(\text{i})$
We have $(2\sin\theta+3\cos\theta)^2+(3\sin\theta-2\cos\theta)^2$
$=4\sin^2\theta+9\cos^2\theta+12\sin\theta\cos\theta\\ \ +9\sin^2\theta+4\cos^2\theta-12\sin\theta\cos\theta$
$=4\big(\sin^2\theta+\cos^2\theta\big)+9\big(\sin^2\theta+\cos^2\theta\big)$
$=4+9$
$=13$
I.e., $(2\sin\theta+3\cos\theta)^2+(3\sin\theta-2\cos\theta)^2=13$
$\Rightarrow2^2+(3\sin\theta-2\cos\theta)^2=13$
$\Rightarrow(3\sin\theta-2\cos\theta)^2=13-4$
$\Rightarrow(3\sin\theta-2\cos\theta)^2=9$
$\Rightarrow(3\sin\theta-2\cos\theta)=\pm3$
View full question & answer→Question 115 Marks
Prove that $\frac{\tan\text{A}}{(1-\cot\text{A})}+\frac{\cot\text{A}}{(1-\tan\text{A})}=(1+\tan\text{A}+\cot\text{A}).$
Answer$\text{LHS}=\frac{\tan\text{A}}{(1-\cot\text{A})}+\frac{\cot\text{A}}{(1-\tan\text{A})}$$=\frac{\tan\text{A}}{(1-\cot\text{A})}+\frac{\cot^2\text{A}}{(\cot\text{A}-1)}$ $\Big[\because\tan\text{A}=\frac{1}{\cot\text{A}}\Big]$
$=\frac{\tan\text{A}}{(1-\cot\text{A})}-\frac{\cot^2\text{A}}{(\cot\text{A}-1)}$
$=\frac{\tan\text{A}-\cot^2\text{A}}{(1-\cot\text{A})}$
$=\frac{\big(\frac{1}{\cot\text{A}}\big)-\cot^2\text{A}}{(1-\cot\text{A})}$
$=\frac{1-\cot^3\text{A}}{\cot\text{A}(1-\cot\text{A})}$
$=\frac{(1-\cot\text{A})(1+\cot\text{A}+\cot^2\text{A})}{\cot\text{A}(1-\cot\text{A})}$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\big]$
$=\frac{1}{\cot\text{A}}+\frac{\cot^2\text{A}}{\cot\text{A}}+\frac{\cot\text{A}}{\cot\text{A}}$
$=1+\tan\text{A}+\cot\text{A}$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 125 Marks
If $5\text{A}=\text{cosec }(\text{A}-36^\circ)$ and $5\text{A}$ is an acute angle, show that $\text{A}=21^\circ.$
AnswerGiven: $\sec\text{5A}=\text{cosec}(\text{A}-36^\circ)$$\Rightarrow\text{cosec}(90^\circ-\text{5A})=\text{cosec}(\text{A}-36^\circ)$ $\big[\because\text{cosec}(90^\circ-\theta)=\sec\theta\big]$
$\Rightarrow90^\circ-\text{5A}=\text{A}-36^\circ$
$\Rightarrow\text{6A}=90^\circ+36^\circ$
$\Rightarrow\text{6A}=126^\circ$
$\Rightarrow\text{A}=21^\circ$
View full question & answer→Question 135 Marks
If $\tan\text{A}=\text{n}\tan\text{B}$ and $\sin\text{A}=\text{m}\sin\text{B},$ prove that $\cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\text{n}^2-1}.$
AnswerWe have, $\tan\text{A}=\text{n}\tan\text{B}$
$\Rightarrow\cot\text{B}=\frac{\text{n}}{\tan\text{A}}\dots(\text{i})$
Again, $\sin\text{A}=\text{m}\sin\text{B}$
$\Rightarrow\text{cosec B}=\frac{\text{m}}{\sin\text{A}}\dots(\text{ii})$
Squaring (i) and (ii) and subtracting (ii) from (i), we get:
$\frac{\text{m}^2}{\sin^2\text{A}}-\frac{\text{n}^2}{\tan^2\text{A}}=\text{cosec}^2\text{B}-\cot^2\text{B}$
$\Rightarrow\frac{\text{m}^2}{\sin^2\text{A}}-\frac{\text{n}^2\cos}{\sin^2\text{A}}=1$
$\Rightarrow\text{m}^2-\text{n}^2\cos^2\text{A}=\sin^2\text{A}$
$\Rightarrow\text{m}^2-\text{n}^2\cos^2\text{A}=1-\cos^2\text{A}$
$\Rightarrow\text{n}^2\cos^2\text{A}-\cos^2\text{A}=\text{m}^2-1$
$\Rightarrow\cos^2\text{A}\big(\text{n}^2-1\big)=\big(\text{m}^2-1\big)$
$\Rightarrow\cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\big(\text{n}^2-1\big)}$
$\therefore\ \cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\big(\text{n}^2-1\big)}$
View full question & answer→Question 145 Marks
Prove the following identities:
$\frac{\sin\theta}{(\sec\theta+\tan\theta)}+\frac{\cos\theta}{(\text{cosec }\theta+\cot\theta-1)}=1$
Answer$\text{LHS}=\frac{\sin\theta}{(\sec\theta+\tan\theta)}+\frac{\cos\theta}{(\text{cosec }\theta+\cot\theta-1)}$ $=\frac{\sin\theta}{\big(\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}-1\big)}+\frac{\cos\theta}{\big(\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}-1\big)}$ $=\frac{\sin\theta\cos\theta}{(1+\sin\theta-\cos\theta)}+\frac{\cos\theta\sin\theta}{(1+\cos\theta-\sin\theta)}$ $=\frac{\sin\theta\cos\theta(1\cos\theta-\sin\theta)+\cos\theta\sin\theta(1+\sin\theta-\cos\theta)}{(1+\sin\theta-\cos\theta)(1+\cos\theta-\sin\theta)}$ $=\frac{\sin\theta\cos\theta+\sin\theta\cos^2\theta-\sin^2\theta\cos\theta+\cos\theta\sin\theta+\cos\theta\sin^2\theta-\cos^2\theta\sin\theta}{(1+\sin\theta-\cos\theta)(1+\cos\theta-\sin\theta)}$ $=\frac{2\sin\theta\cos\theta}{1+\cos\theta-\sin\theta+\sin\theta\cos\theta-\sin^2\theta-\cos\theta-\cos^2\theta+\cos\theta\sin\theta}$ $=\frac{2\sin\theta\cos\theta}{2\sin\theta\cos\theta}=1$ $=\text{R.H.S.}$ $\therefore\ \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 155 Marks
If $(\sin\theta+\cos\theta)=\sqrt{2}\cos\theta,$ show that $\cot\theta=\big(\sqrt{2}+1\big).$
AnswerWe have,
$(\sin\theta+\cos\theta)=\sqrt{2}\cos\theta$
Dividing both side by $\sin\theta,$ we get:
$\frac{\sin\theta}{\sin\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sqrt{2}\cos\theta}{\sin\theta}$
$\Rightarrow1+\cot\theta=\sqrt{2}\cot\theta$
$\Rightarrow\sqrt{2}\cot\theta-\cot\theta=1$
$\Rightarrow\Big(\sqrt{2}-1\Big)\cot\theta=1$
$\Rightarrow\cot\theta=\frac{1}{\big(\sqrt{2}-1\big)}$
$\Rightarrow\cot\theta=\frac{1}{\big(\sqrt{2}-1\big)}\times\frac{\big(\sqrt{2}+1\big)}{\big(\sqrt{2}+1\big)}$
$\Rightarrow\cot\theta=\frac{\big(\sqrt{2}+1\big)}{2-1}$
$\Rightarrow\cot\theta=\frac{\big(\sqrt{2}+1\big)}{1}$
$\therefore\ \cot\theta=\big(\sqrt{2}+1\big)$
View full question & answer→Question 165 Marks
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$ and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n},$ prove that $\big(\text{m}^2+\text{n}^2\big)=\big(\text{a}^2+\text{b}^2\big).$
AnswerWe have $\text{m}^2+\text{n}^2=\big[(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2\big]$
$=\big(\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\cos\theta\sin\theta\big)\\ \ +\big(\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta\big)$
$=\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta$
$=\big(\text{a}^2\cos^2\theta+\text{a}^2\sin^2\theta\big)+\big(\text{b}^2\cos^2\theta+\text{b}^2\sin^2\theta\big)$
$=\text{a}^2\big(\cos^2\theta+\sin^2\theta\big)+\text{b}^2\big(\cos^2\theta+\sin^2\theta\big)$
$=\text{a}^2+\text{b}^2$ $\big[\because\ \sin^2+\cos^2=1\big]$
Hence, $\text{m}^2+\text{n}^2=\text{a}^2+\text{b}^2$
View full question & answer→Question 175 Marks
Prove the following identities:
$\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}=(\sec\theta+\tan\theta)^2$
$=1+2\tan^2\theta+2\sec\theta\tan\theta$
Answer$\text{LHS}=\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}\times\frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}$
$=\frac{(\sec\theta+\tan\theta)^2}{\big(\sec^2\theta-\tan^2\theta\big)}=(\sec\theta+\tan\theta)^2$
Further,
$(\sec\theta+\tan\theta)^2$
$=\sec^2\theta+\tan^2\theta+2\sec\theta\tan\theta$
$=1+\tan^2\theta+\tan^2\theta+2\sec\theta\tan\theta$
$=1+2\tan^2+2\sec\theta\tan\theta$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 185 Marks
If $(\sec\theta+\tan\theta)=\text{m}$ and $(\sec\theta-\tan\theta)=\text{n},$ show that mn = 1.
AnswerWe have $(\sec\theta+\tan\theta)=\text{m}\dots(\text{i})$
Again, $(\sec\theta-\tan\theta)=\text{n}\dots(\text{ii})$
Now, multiplying (i) and (ii), we get:
$(\sec\theta+\tan\theta)\times(\sec\theta-\tan\theta)=\text{mn}$
$\Rightarrow\sec^2\theta-\tan^2\theta=\text{mn}$
$\Rightarrow1=\text{mn}$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$\therefore\ \text{mn}=1$
View full question & answer→Question 195 Marks
If $\text{x}=\text{a}\sec\theta+\text{b}\tan\theta$ and $\text{y}=\text{a}\tan\theta-\text{b}\sec\theta,$ prove that $\big(\text{x}^2+\text{y}^2\big)=\big(\text{a}^2-\text{b}^2\big).$
AnswerWe have $\text{x}^2+\text{y}^2=\big[(\text{a}\sec\theta+\text{b}\tan\theta)^2+(\text{a}\tan\theta-\text{b}\sec\theta)^2\big]$
$=\big(\text{a}^2\sec^2\theta+\text{b}^2\tan^2\theta+2\text{ab}\sec\theta\tan\theta\big)\\ \ -\big(\text{a}^2\tan^2\theta+\text{b}^2\sec^2\theta+2\text{ab}\tan\theta\sec\theta\big)$
$=\text{a}^2\sec^2\theta+\text{b}^2\tan^2\theta-\text{a}^2\tan^2\theta+\text{b}^2\sec^2\theta$
$=\big(\text{a}^2\sec^2\theta+\text{a}^2\tan^2\theta\big)-\big(\text{b}^2\sec^2\theta+\text{b}^2\tan^2\theta\big)$
$=\text{a}^2\big(\sec^2\theta-\tan^2\theta\big)-\text{b}^2\big(\sec^2\theta-\tan^2\theta\big)$
$=\text{a}^2-\text{b}^2$ $\big[\because\ \sec^2+\tan^2=1\big]$
Hence, $\text{x}^2-\text{y}^2=\text{a}^2-\text{b}^2$
View full question & answer→Question 205 Marks
If $\text{m}=(\cos\theta-\sin\theta)$ and $\text{n}=(\cos\theta+\sin\theta),$ then show that $\sqrt{\frac{\text{m}}{\text{n}}}+\sqrt{\frac{\text{n}}{\text{m}}}=\frac{2}{\sqrt{1-\tan^2\theta}}.$
Answer$\text{LHS}=\sqrt{\frac{\text{m}}{\text{n}}}+\sqrt{\frac{\text{n}}{\text{m}}}$$=\sqrt{\frac{\text{m}}{\text{n}}}+\sqrt{\frac{\text{n}}{\text{m}}}$
$=\frac{\text{m}+\text{n}}{\sqrt{\text{mn}}}$
$=\frac{(\cos\theta-\sin\theta)+(\cos\theta+\sin\theta)}{\sqrt{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}}$
$=\frac{2\cos\theta}{\sqrt{\cos^2\theta-\sin^2\theta}}$
$=\frac{\big(\frac{2\cos\theta}{\cos\theta}\big)}{\bigg(\frac{\sqrt{cos^2\theta-\sin^2\theta}}{\cos\theta}\bigg)}$
$=\frac{2}{\sqrt{\frac{\cos^2\theta}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}}}$
$=\frac{2}{1-\tan^2\theta}$
$=\text{RHS}$
View full question & answer→Question 215 Marks
If $\text{x}=\text{a}\cos^3\theta$ and $\text{y}=\text{b}\sin^3\theta,$ prove that $\Big(\frac{\text{x}}{\text{a}}\Big)^\frac{2}{3}+\Big(\frac{\text{y}}{\text{b}}\Big)^\frac23=1.$
AnswerWe have $\text{x}=\text{a}\cos^3\theta$
$\Rightarrow\frac{\text{x}}{\text{a}}=\cos^3\theta\dots(\text{i})$
Again, $\text{y}=\text{b}\sin^3\theta$
$\Rightarrow\frac{\text{y}}{\text{b}}=\sin^3\theta\dots(\text{ii})$
Now, $\text{LHS}=\Big(\frac{\text{x}}{\text{a}}\Big)^\frac23+\Big(\frac{\text{y}}{\text{b}}\Big)^\frac{2}{3}$
$=\big(\cos^3\theta\big)^\frac23+\big(\sin^3\theta\big)^\frac23$ [From (i) and (ii)]
$=\cos^2\theta+\sin^2\theta$
$=1$
Hence, LHS = RHS.
View full question & answer→Question 225 Marks
If $(\cot\theta+\tan\theta)=\text{m}$ and $(\sec\theta-\cos\theta)=\text{n},$ prove that $\big(\text{m}^2\text{n}\big)^\frac23-\big(\text{m}\text{n}^2\big)^\frac23=1.$
AnswerWe have $(\cot\theta+\tan\theta)=\text{m}$ and $(\sec\theta-\cos\theta)=\text{n}$
Now, $\text{m}^2\text{n}=\Big[\big(\cot\theta+\tan\theta\big)^2\big(\sec\theta-\cos\theta\big)\Big]$
$=\Big[\Big(\frac{1}{\tan\theta}+\tan\theta\Big)^2\Big(\frac{1}{\cos\theta}-\cos\theta\Big)\Big]$
$=\frac{\big(1+\tan^2\theta\big)^2}{\tan^2\theta}\times\frac{\big(1-\cos^2\theta\big)}{\cos\theta}$
$=\frac{\sec^4\theta}{\tan^2\theta}\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\sec^4\theta}{\frac{\sin^2\theta}{\cos^2\theta}}\times\frac{\sin^2\theta}{\cos\theta}$
$=\frac{\cos^2\theta\times\sec^4\theta}{\cos\theta}$
$=\cos\theta\sec^4\theta$
$=\frac{1}{\sec\theta}\times\sec^4\theta=\sec^3\theta$
$\therefore\ \big(\text{m}^2\text{n}\big)^\frac23=\big(\sec^3\theta\big)^\frac23=\sec^2\theta$
Again, $\text{mn}^2=\Big[(\cot\theta+\tan\theta)(\sec\theta-\cos\theta)^2\Big]$
$=\bigg[\Big(\frac{1}{\tan\theta}+\tan\theta\Big).\Big(\frac{1}{\cos\theta}-\cos\theta\Big)^2\bigg]$
$=\frac{\big(1+\tan^2\theta\big)}{\tan\theta}\times\frac{\big(1-\cos^2\theta\big)^2}{\cos^2\theta}$
$=\frac{\sec^2\theta}{\tan\theta}\times\frac{\sin^4\theta}{\cos^2\theta}$
$=\frac{\sec^2\theta}{\frac{\sin\theta}{\cos\theta}}\times\frac{\sin^4\theta}{\cos^2\theta}$
$=\frac{\sec^2\theta\times\sin^3\theta}{\cos\theta}$
$=\frac{1}{\cos^2\theta}\times\frac{\sec^3\theta}{\cos\theta}=\tan^3\theta$
$\therefore\ \big(\text{m}\text{n}^2\big)^\frac23=\big(\tan^3\theta\big)^\frac23=\tan^2\theta$
Now, $\big(\text{m}^2\text{n}\big)^\frac23-\big(\text{m}\text{n}^2\big)^\frac23$
$=\sec^2\theta-\tan^2\theta=1$
$=\text{RHS}$
Hence proved.
View full question & answer→Question 235 Marks
Prove the following identities:
$\frac{(\sin\text{A}-\sin\text{B})}{(\cos\text{A}+\cos\text{B})}+\frac{(\cos\text{A}-\cos\text{B})}{(\sin\text{A}+\sin\text{B})}=0$
Answer$\text{LHS}=\frac{(\sin\text{A}-\sin\text{B})}{(\cos\text{A}+\cos\text{B})}+\frac{(\cos\text{A}-\cos\text{B})}{(\sin\text{A}+\sin\text{B})}$
$=\frac{(\sin\text{A}+\sin\text{B})(\sin\text{A}-\sin\text{B})+(\cos\text{A}+\cos\text{B})(\cos\text{A}-\cos\text{B})}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{\sin^2\text{A}-\sin^2\text{B}+\cos^2\text{A}-\cos^2\text{B}}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{\big(\sin^2\text{A}+\cos^2\text{A}\big)-\big(\sin^2\text{A}+\cos^2\text{B}\big)}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{1-1}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=0$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 245 Marks
Prove the following identities:
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\frac{2}{\big(\sin^2\theta-\cos^2\theta\big)}=\frac{2}{\big(2\sin^2\theta-1\big)}$
Answer$\text{LHS}=\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2}{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}$
$=\frac{\big(\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta\big)+\big(\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta\big)}{\sin^2\theta-\cos^2\theta}$
$=\frac{(1+2\sin\theta\cos\theta)+(1-2\sin\theta\cos\theta)}{\sin^2\theta-\cos^2\theta}$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\frac{2}{\sin^2\theta-\cos^2\theta}$
Also, $\frac{2}{\sin^2\theta-\cos^2\theta}=\frac{2}{\sin^2\theta-\big(1-\sin^2\theta\big)}$
$=\frac{2}{2\sin^2\theta-1}$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 255 Marks
Prove the following identities:
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\frac{2}{\big(1-2\cos^2\theta\big)}$
Answer$\text{LHS}=\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2}{\sin^2\theta-\cos^2\theta}$
$=\frac{\sin^2\theta+\cos^2\theta+2\cos\theta\sin\theta+\sin^2\theta+\cos^2\theta+2\cos\theta\sin\theta}{1-\cos^2\theta-\cos^2\theta}$
$=\frac{1+1}{1-2\cos^2\theta}$
$=\frac{2}{\big(1-2\cos^2\theta\big)}$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
View full question & answer→Question 265 Marks
If $\sec\theta+\tan\theta=\text{p},$ prove that:$\sin\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$
AnswerWe have,
$\sec\theta+\tan\theta=\text{p}\dots(1)$
$\Rightarrow\frac{\sec\theta+\tan\theta}{1}\times\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{1}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{p}}\dots(2)$
Adding (1) and (2), we get:
$2\sec\theta=\Big(\text{p}+\frac{1}{\text{p}}\Big)$
$\Rightarrow\sec\theta=\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)\dots(\text{A})$
Subtracting (2) from (1), we get:
$2\tan\theta=\Big(\text{p}-\frac{1}{\text{p}}\Big)$
$\Rightarrow\tan\theta=\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)\dots(\text{B})$
Using (A) and (B), we get:
$\sin\theta=\frac{\tan\theta}{\sec\theta}$
$=\frac{\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)}{\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)}$
$=\frac{\Big(\frac{\text{p}^2-1}{\text{p}}\Big)}{\Big(\frac{\text{p}^2+1}{\text{p}}\Big)}$
$\therefore\ \sin\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$
View full question & answer→Question 275 Marks
Prove that $\frac{\sin\text{A}-2\sin^3\text{A}}{\big(2\cos^2\text{A}-\cos\text{A}\big)}=\tan\text{A}.$
Answer$\text{LHS}=\frac{\big(\sin\text{A}-2\sin^2\text{A}\big)}{\Big(2\cos^2\text{A}-\cos\text{A}\big)}$$=\frac{\sin\text{A}\big(1-2\sin^2\text{A}\big)}{\cos\text{A}\big(2\cos^2\text{A}-1\big)}$
$=\tan\text{A}\Bigg\{\frac{\big(\sin^2\text{A}+\cos^2\text{A}-2\sin^2\text{A}\big)}{2\cos^2\text{A}-\sin^2\text{A}-\cos^2\text{A}}\Bigg\}$ $\big[\because\sin^2\text{A}+\cos^2\text{A}=1\big]$
$=\tan\text{A}\Bigg\{\frac{\big(\cos^2\text{A}-\sin^2\text{A}\big)}{\big(\cos^2\text{A}-\sin^2\text{A}\big)}\Bigg\}$
$=\tan\text{A}$
$=\text{RHS}$
View full question & answer→Question 285 Marks
Prove that $\frac{1}{(sec\theta-\tan\theta)}-\frac{1}{\cos\theta}=\frac{1}{\cos\theta}-\frac{1}{(\sec\theta+\tan\theta)}.$
Answer$\frac{1}{(\sec\theta-\tan\theta)}-\frac{1}{\cos\theta}=\frac{1}{\cos\theta}-\frac{1}{(\sec\theta+\tan\theta)}$$\text{LHS}=\frac{1}{\sec\theta-\tan\theta}-\frac{1}{\cos\theta}$
$=\frac{(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}-\sec\theta$ $($Multiplying the numerator and denominator by $\sec\theta+\tan\theta)$
$=\frac{\sec\theta+\tan\theta}{\sec^2\theta-\tan^2\theta}-\sec\theta$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\text{RHS}=\frac{1}{\cos\theta}-\frac{1}{\sec\theta+\tan\theta}$
$=\sec\theta-\frac{(\sec\theta-\tan\theta)}{\sec^2\theta-\tan^2\theta}$ $($Multiplying the numerator and denominator by $\sec\theta-\tan\theta)$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.
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