Question 13 Marks
Find the values of the other five trigonometric functions in the following:
$\sin\text{x}=\frac{3}{5},$ x in quadrant I
AnswerWe have: $\sin\text{x}=\frac{3}{5}$ and x are in the first quadrant.In the first quadrant, all six T -ratios are positive.$\therefore\cos\text{x}=\sqrt{1-\sin^2\text{x}}=\sqrt{1-\Big(\frac{3}{5}\Big)^2}=\frac{4}{5}$
$\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
$\cot\text{x}=\frac{1}{\tan\text{x}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
$\sec\text{x}=\frac{1}{\cos\text{x}}=\frac{1}{\frac{4}{5}}=\frac{5}{4}$
$\text{cosec}\text{ x}=\frac{1}{\sin\text{x}}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$
View full question & answer→Question 23 Marks
Prove that:
$3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}=1$
Answer$\text{L.H.S}=3\sin\frac{\pi}{6}\sec\frac{\pi}{3}-4\sin\frac{5\pi}{6}\cot\frac{\pi}{4}$
$=3\times\frac{1}{2}\times2-4\sin\Big(\pi-\frac{\pi}{6}\Big)\times1$
$=3-4\sin\frac{\pi}{6}$
$=3-4\times\frac{1}{2}$
$= 3 - 2$
$= 1$
$\text{= R.H.S}$
$\text{Proved}$
View full question & answer→Question 33 Marks
Prove the following identities:
$1-\frac{\sin^2\text{x}}{1+\cot\text{x}}-\frac{\cos^2\text{x}}{1+\tan\text{x}}=\sin\text{x}\cos\text{x}$
Answer$\text{L.H.S}=1-\frac{\sin^2\text{x}}{1+\cot\text{x}}-\frac{\cos^2\text{x}}{1+\tan\text{x}}$
$=1-\frac{\sin^2\text{x}}{1+\frac{\cos\text{x}}{\sin\text{x}}}-\frac{\cos^2\text{x}}{1+\frac{\sin\text{x}}{\cos\text{x}}}$ $\Big(\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}},\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}\Big)$
$=1-\frac{\sin^2\text{x}}{\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}}-\frac{\cos^2\text{x}}{\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}}}$
$=1-\frac{\sin^3\text{x}}{\sin\text{x}+\cos\text{x}}-\frac{\cos^3\text{x}}{\cos\text{x}+\sin\text{x}}$
$=\frac{\sin\text{x}+\cos\text{x}-(\sin^3+\cos^3\text{x})}{\sin\text{x}+\cos\text{x}}$
$=\frac{\sin\text{x}+\cos\text{x}-(\sin\text{x}+\cos\text{x})(\sin^3\text{x}+\cos^2\text{x}-\sin\text{x}\cos\text{x})}{\sin\text{x}+\cos\text{x}}$ $(\text{Using a}^3+\text{b}^3=(\text{a+b})(\text{a}^2+\text{b}^2-\text{ab}))$
$\frac{(\sin\text{x}+\cos\text{x})(1-(1-\sin\text{x}\cos\text{x}))}{\sin\text{x}+\cos\text{x}}$ $(\text{Using }\sin^2\text{x}+\cos^2\text{x}=1)$
$=\sin\text{x}\cos\text{x}$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 43 Marks
If $\tan\text{x}=\frac{\text{a}}{\text{b}},$ show that $\frac{\text{x}\sin\text{x - b}\cos\text{x}}{\text{a}\sin\text{x}+\text{b}\cos\text{x}}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}.$
AnswerGiven $=\tan\text{x}=\frac{\text{a}}{\text{b}}$
To show: $\frac{\text{a}\sin\text{x}-\text{b}\cos\text{x}}{\text{a}\sin\text{x}+\text{b}\cos\text{x}}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$.
Since, $\tan\text{x}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\frac{\sin\text{x}}{\cos\text{x}}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\text{b}\sin\text{x}=\text{a}\cos\text{x}=\lambda$ (Say)
$\Rightarrow\sin\text{x}=\frac{\lambda}{\text{b}}$ and $\cos\text{x}=\frac{\lambda}{\text{a}}$
How $\frac{\text{a}\sin\text{x}-\text{b}\cos\text{x}}{\text{a}\sin\text{x}+\text{b}\cos\text{x}}=\frac{\frac{\text{a}.\lambda}{\text{b}}-\frac{\text{b}.\lambda}{\text{a}}}{\frac{\text{a}.\lambda}{\text{b}}+\frac{\text{b}.\lambda}{\text{a}}}$
$=\frac{\lambda\Big(\frac{\text{a}}{\text{b}}-\frac{\text{b}}{\text{a}}\Big)}{\lambda\Big(\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}\Big)}$
$=\frac{\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}}{\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}}$
$=\frac{\frac{\text{a}^2-\text{b}^2}{\text{ab}}}{\frac{\text{a}^2+\text{b}^2}{\text{ab}}}$
$=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
$\text{Proved}$
View full question & answer→Question 53 Marks
Prove that:
$\frac{\sin(\pi+\text{x})\cos\big(\frac{\pi}{2}+\text{x}\big)\tan\big(\frac{3\pi}{2}-\text{x}\big)\cot(2\pi-\text{x})}{\sin(2\pi-\text{x})\cos(2\pi+\text{x})\text{cosec}(-\text{x})\sin\big(\frac{3\pi}{2}-\text{x}\big)}=1$
Answer$\text{L.H.S}=\frac{\sin(\pi+\text{x})\cos\big(\frac{\pi}{2}+\text{x}\big)\tan\big(\frac{3\pi}{2}-\text{x}\big)\cot(2\pi-\text{x})}{\sin(2\pi-\text{x})\cos(2\pi+\text{x})\text{cosec}(-\text{x})\sin\big(\frac{3\pi}{2}-\text{x}\big)}$
$=\frac{\sin\text{x}(-\sin\text{x})\cot(-\cot\text{x})}{-\sin\text{x}\cos\text{x}(-\text{cosec})(-\cos\text{x})}$ $\begin{pmatrix}\because\tan(270^\circ-\text{x})=\cot\text{x}\\\&\sin(270^\circ=\text{x})=-\cos\text{x}\end{pmatrix}$
$=\frac{-\sin\text{x}\times\sin\text{x}\times\cos\text{x}\times\cos\text{x}\times\sin\text{x}}{-\sin\text{x}\times\cos\text{x}\times\sin\text{x}\times\sin\text{x}\text{x}\cos\text{x}}$ $\begin{pmatrix}\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\\\&\text{ cosec}\text{x}=\frac{1}{\sin\text{x}}\end{pmatrix}$
$= 1$
$\text{= R.H.S}$
$\text{Proved}$
View full question & answer→Question 63 Marks
Prove that:
$\sin\frac{8\pi}{3}\cos\frac{23\pi}{6}+\cos\frac{13\pi}{3}\sin\frac{35\pi}{6}=\frac{1}{2}$
Answer$\text{L.H.S}=\sin\frac{8\pi}{3}\cos\frac{23\pi}{6}+\cos\frac{13\pi}{3}\sin\frac{35\pi}{6}$
$=\sin\Big(3\pi-\frac{\pi}{3}\Big)\cos\Big(4\pi-\frac{\pi}{6}\Big)+\cos\Big(4\pi+\frac{\pi}{3}\Big)\sin\Big(6\pi-\frac{\pi}{6}\Big)$
$=\sin\frac{\pi}{3}\cos\frac{\pi}{6}+\cos\frac{\pi}{3}\Big(-\sin\frac{\pi}{6}\Big)$ $(\because\sin(6\pi-\theta)=-\sin\theta)$
$=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{22}+\frac{1}{2}\times\Big(\frac{-1}{2}\Big)$
$=\frac{3}{2}-\frac{1}{4}$
$=\frac{2}{4}$
$=\frac{1}{2}$
$\text{= R.H.S}$
$\text{Proved}$
View full question & answer→Question 73 Marks
Prove the following identities:
$\frac{\tan\text{x}}{1-\cot\text{x}}+\frac{\cot\text{x}}{1-\tan\text{x}}=(\sec\text{x}\text{ cosec x}+1)$
Answer$\text{L.H.S}=\frac{\tan\text{x}}{1-\cot\text{x}}+\frac{\cot\text{x}}{1-\tan\text{x}}$
$=\frac{\big(\frac{\sin\text{x}}{\cos\text{x}}\big)}{\big(1-\frac{\cos\text{x}}{\sin\text{x}}\big)}+\frac{\big(\frac{\cos\text{x}}{\sin\text{x}}\big)}{1-\frac{\sin\text{x}}{\cos\text{x}}}$
$=\frac{\sin\text{x}}{\cos\text{x}\frac{\big(\sin\text{x}-\cos\text{x}\big)}{\sin\text{x}}}+\frac{\cos\text{x}}{\sin\text{x}\frac{\big(\cos\text{x}-\sin\text{x}\big)}{\cos\text{x}}}$
$=\frac{\sin^{2}\text{x}}{\cos\text{x}\big(\sin\text{x}-\cos\text{x}\big)}+\frac{\cos^{2}\text{x}}{\sin\text{x}\big(\cos\text{x}-\sin\text{x}\big)}$
$=\frac{\sin^{3}\text{x}-\cos^{3}\text{x}}{\cos\text{x}\sin\text{x}\big(\sin\text{x}-\cos\text{x}\big)}$
$=\frac{\big(\sin\text{x}-\cos\text{x}\big)\big(\sin^{2}\text{x}+\cos^{2}\text{x}+\sin\text{x}\cos\text{x}\big)}{\cos\text{x}\sin\text{x}\big(\sin\text{x}-\cos\text{x}\big)}$ $ \big[\text{using }\text{a}^{3}-\text{b}^{3}=\big(\text{a}-\text{b}\big)\big(\text{a}^{2}+\text{b}^{2}+\text{ab}\big)\big]$
$=\frac{1+\sin\text{x}\cos\text{x}}{\sin\text{x}\cos\text{x}}$ $ \big[\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\big]$
$=\frac{1}{\sin\text{x}\cos\text{x}}+\frac{\sin\text{x}\cos\text{x}}{\sin\text{x}\cos\text{x}}$
$=\sec\text{x}\text{ cosec }\text{x}+1$ $\big[\because\frac{1}{\cos\text{x}}=\sec\text{x},\frac{1}{\sin\text{x}}=\text{cosec }\text{x}\big]$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 83 Marks
Find the values of the other five trigonometric functions in the following:
$\cot\text{x}=\frac{12}{5},$ x in quadrant III.
AnswerWe have, $\cot\text{x} = \frac{12}{5}$ and x are in the third quardrant.In the third quardrant, $\tan\text{x}$ and $\cot\text{x}$ are positive And,$$ $\sin\text{x},\cos\text{x},\sec\text{x}$ and $\text{cosec}\text{ x}$ are negative.$\therefore\tan\text{x}=\frac{1}{\cot\text{x}}=\frac{1}{\frac{12}{5}}=\frac{5}{12}$
$\text{cosec}\text{x}=-\sqrt{1+\cot^2\text{x}}$
$=-\sqrt{1+(\frac{12}{5})^2}=-\frac{13}{5}$
$\sin\text{x}=\frac{1}{\text{cosec}}=\frac{1}{-\frac{13}{5}}=-\frac{5}{13}$
$\cos\text{x}=-\sqrt{1-\sin^2\text{x}}$
$=-\sqrt{1-\Big(\frac{-5}{13}\Big)^2}=-\frac{13}{5}$
And, $\sec\text{x}=\frac{1}{\cos\text{x}}=\frac{1}{\frac{-12}{13}}=\frac{-13}{12}$
View full question & answer→Question 93 Marks
If $\sin\text{x}=\frac{3}{5},\tan\text{y}=\frac{1}{2}$ and $\frac{\pi}{2}<\text{x}<\pi<\text{y}<\frac{3\pi}{2},$ find the value of $8\tan\text{x}-\sqrt{5}\sec\text{y}.$
AnswerWe have$\sin\text{x}=\frac{3}{5},\tan\text{y}=\frac{1}{2}$and$\frac{\pi}{2}<\text{x}<\pi<\text{y}<\frac{3\pi}{2}$,
Thus, x is in the second quadrant and y is in the third quadrant. In the second quadrant, $\cos\text{x}$ and $\tan\text{x}$ are negative. In the third quadrant, $\sec\text{y}$ is negative.$\therefore\cos\text{x} = -\sqrt{1-\sin^2\text{x}} =-\sqrt{1-\Big(\frac{3}{5}\Big)^2} =\frac{-4}{5}$
$\tan\text{x} =\frac{\frac{3}{5}}{\frac{-4}{5}} =\frac{-3}{4}$
And, $\sec\text{y} =-\sqrt{1+\tan^2\text{y}} = -\sqrt{1+\Big(\frac{1}{2}\Big)^2} = \frac{\sqrt{-5}}{2}$$\therefore 8 \tan\text{x} - \sqrt5 \sec\text{y} = 8 \times \frac{-3}{4} - \sqrt5 \times \frac{-\sqrt5}{2}= -6+\frac{5}{2}=\frac{7}{2}$
View full question & answer→Question 103 Marks
If $\cot \text{x}(1+\sin\text{x})=4\text{ m}$ and $\cot \text{x}(1-\sin\text{x})=4\text{ n},$prove that $\text{(m}^2-\text{n}^2)^2=\text{mn.}$
AnswerLet, $\cot\text{x}(1+\sin\text{x})=4\text{m}\cdots\text{(i)}$
and, $\cot\text{x}(1-\sin\text{x})=4\text{n}\cdots\text{(ii)}$
To show: $\text{(m}^2-\text{n}^2)^2=\text{mn}$
From (i) and (ii), we get
$\text{m}=\frac{\cot\text{x}(1+\sin\text{x})}{4}\&\text{ n}=\frac{\cot\text{x}(1-\sin\text{x})}{4}$
$\text{L.H.S}=(\text{m}^2-\text{n}^2)^2$
$=((\text{m}+\text{n})(\text{m}-\text{n}))^2$
$=(\text{m}+\text{n})^2(\text{m}-\text{n})^2$
$=\Big(\frac{\cot\text{x}(1+\sin\text{x})+\cot\text{x}(1-\sin\text{x})}{4}\Big)^2\times\Big(\frac{\cot\text{x}(1+\sin\text{x})-\cot\text{x}(1-\sin\text{x})}{4}\Big)^2$
$=\Big(\frac{\cot\text{x}(1+\sin\text{x}+1-\sin\text{x})}{4}\Big)\times\Big(\frac{\cot\text{x}(1+\sin\text{x}-1-\sin\text{x})}{4}\Big)^2$
$=\frac{\cot^2\text{x}}{16}\times4\times\frac{\cot^2\text{x}}{16}\times\sin^2\text{x}$
$\frac{\cot^2\text{x}}{16}\times\frac{\cos^2\text{x}}{\sin^2\text{x}}\sin^2\text{x}$ $\Big[\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\Big]$
$=\frac{\cot\text{x}}{4}\times\frac{\cot\text{x}}{4}\times(1-\sin^2\text{x})$ $\Big[\because\cos^2\text{x}=1=\sin^2\text{x}\Big]$
$=\frac{\cot\text{x}(1+\sin\text{x})}{4}\times\frac{\cot\text{x}(1-\sin\text{x})}{4}$
$=\text{mn}.$
View full question & answer→Question 113 Marks
Prove the following identities:
$(\text{cosec}\text{x}−\sin\text{x})(\sec\text{x}−\cos\text{x})(\tan\text{x}+\cot\text{x})=1$
Answer$\text{L.H.S}=\big(\text{cosec }\text{x}-\sin\text{x}\big)\big(\sec\text{x}-\cos\text{x}\big)\big(\tan\text{x}+\cot\text{x}\big)$
$=\Big(\frac{1}{\sin\text{x}}-\sin\text{x}\Big)\Big(\frac{1}{\cos\text{x}}-\cos\text{x}\Big)\Big(\frac{\sin\text{x}}{\cos\text{x}}+\frac{\cos\text{x}}{\sin\text{x}}\Big)$$\begin{bmatrix}\because\text{cosec}=\frac{1}{\sin\text{x}},\sec\text{x}=\frac{1}{\cos\text{x}},\\\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}} \text{and}\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\end{bmatrix}$
$=\Big(\frac{1-\sin^{2}\text{x}}{\sin\text{x}}\Big)\Big(\frac{1-\cos^{2}\text{x}}{\cos\text{x}}\Big)\Big(\frac{\sin\text{x}}{\cos\text{x}}+\frac{\cos\text{x}}{\sin\text{x}}\Big)$
$=\frac{\cos^{2}\text{x}.\sin^{2}\text{x}.1}{\sin^{2}\text{x}.\cos^{2}\text{x}}$ $\begin{pmatrix}\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\\\Rightarrow1-\sin^{2}\text{x}=\cos^{2}\text{x},\text{and}1-\cos^{2}\text{x}=\sin^{2}\text{x}\end{pmatrix}$
$=1$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 123 Marks
Prove that : $\frac{\text{cosec}(90^{\circ}+\text{x})+\cot(450^\circ+\text{x})}{\text{cosec}(90^\circ-\text{x})+\tan(180^\circ-\text{x})} +\frac{\tan\text{x}(180^\circ+\text{x})+\sec(180^\circ-\text{x})}{\tan(360^\circ+\text{x})-\sec(-\text{x})}=2$
Answer$\text{L.H.S}=\frac{\text{cosec}(90^{\circ}+\text{x})+\cot(450^\circ+\text{x})}{\text{cosec}(90^\circ-\text{x})+\tan(180^\circ-\text{x})}+\frac{\tan\text{x}(180^\circ+\text{x})+\sec(180^\circ-\text{x})}{\tan(360^\circ+\text{x})-\sec(-\text{x})}$
$=\frac{\sec\text{x}+\cot\Big(2\pi+\frac{\pi}{2}+\text{x}\Big)}{\sec\text{x}-\tan\text{x}}+\frac{\tan\text{x}-\sec\text{x}}{\tan\text{x}-\sec\text{x}}$
$\left(\begin{array}{c}\because\text{cosec}(90^\circ+\text{x})=\sec\text{x},\text{cosec}(90^\circ+\text{x})=\sec\text{x},\tan(180^\circ-\text{x})\\=-\tan\text{x}\sec(-\text{x})=\sec\text{x}\end{array}\right)$
$=\frac{\sec\text{x}+\cot\Big(\frac{\pi}{2}+\text{x}\Big)}{\sec\text{x}-\tan\text{x}}+1$ $(\because\cot(2\pi+\text{x})=\cot\text{x})$
$=\frac{\sec\text{x}-\tan\text{x}}{\sec\text{x}-\tan\text{x}}+1$ $\Big(\because\cot\Big(\frac{\pi}{2}+\text{x}\Big)=-\tan\text{x}\Big)$
$= 1 + 1$
$= 2$
$= \text{R.H.S}$ Proved
View full question & answer→Question 133 Marks
Prove the following identities:
$\sin^6\text{x}+\cos^6\text{x}=1−3\sin^2\text{x}\cos^2\text{x}$
Answer$\text{L.H.S}=\sin^{6}\text{x}+\cos^{6}\text{x}$
$=\big(\sin^{2}\text{x}\big)^{3}+\big(\cos^{6}\text{x}\big)$
$=\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)$ $\bigg[\big(\sin^{2}\text{x}\big)^{2}-\big(\sin^{2}\text{x}\big)-\sin^{2}\text{x}\cos^{2}\text{x}+\big(\cos^{2}\text{x}\big)\bigg]$ $\big(\because\text{a}^{3}+\text{b}^{3}=\big(\text{a}+\text{b}\big)\big(\text{a}^{2}-\text{ab}+\text{b}^{2}\big)$
$=\big(\sin\text{x}\big)^{2}+\big(\cos\text{x}\big)^{2}$ $\Big[\big(\sin^{2}\text{x}\big)^{2}+2\sin^{2}\text{x}\cos^{2}\text{x}-2\sin^{2}\text{x}\cos^{2}\text{x}-\sin^{2}\cos^{2}\text{x}\big)\Big]$
$=\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)^{2}-3\sin^{2}\text{x}\cos^{2}\text{x}$
$=\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)^{2}-3\sin^{2}\text{x}\cos^{2}\text{x}$
$=1^{2}-3\sin^{2}\text{x}\cos^{2}\text{x}$ $\big(\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\big)$
$=1-3\sin^{2}\text{x}\cos^{2}\text{x}$
$=\text{R.H.S}$
$\text{L.H.S = R.H.S}$
View full question & answer→Question 143 Marks
Find the values of the other five trigonometric functions in the following:
$\tan\text{x}=-\frac{3}{4},$ x in quadrant III
AnswerWe have: $\tan\text{x}=\frac{3}{4}$ and x are in the third quadrant. In the third quadrant, $\tan\text{x},$ and $\cot\text{x}$ are positive and $\sin\text{x},\cos\text{x},\sec\text{x}$ and $\text{cosec}\text{ x}$ are negative.$\therefore\cot\text{x}=\frac{1}{\tan\text{x}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
$\sec\text{x}=-\sqrt{1+\tan^2\text{x}}=-\sqrt{1+\Big(\frac{3}{4}\Big)^2}=-\frac{5}{4}$
$\cos\text{x}=\frac{1}{\sec\text{x}}=\frac{1}{-\frac{5}{4}}=-\frac{4}{5}$
$\sin\text{x}=-\sqrt{1-\cos^2\text{x}}=-\sqrt{1-\Big(\frac{-4}{5}\Big)^2}=\frac{-3}{5}$
$\text{cosec}\text{ x}=\frac{1}{\sin\text{x}}=\frac{1}{-\frac{3}{5}}=-\frac{5}{3}$
View full question & answer→Question 153 Marks
If $\sin\text{x}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$ find the value of $\tan\text{x},\sec\text{x}$ and $\text{cosec x}$
AnswerNow, $\cos\text{x}=\sqrt{1-\sin^2\text{x}}$ $=\sqrt{\frac{1-(\text{a}^2-\text{b}^2)}{(\text{a}^2+\text{b}^2)^2}}$ $\Big[\because\sin\text{x}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}\Big]$$=\sqrt{\frac{(\text{a}^2+\text{b}^2)^2-(\text{a}^2-\text{b}^2)^2}{(\text{a}^2+\text{b}^2)^2}}$
$=\sqrt{\frac{(\text{a}^2+\text{b}^2+\text{a}^2-\text{b}^2)(\text{a}^2+\text{b}^2-\text{a}^2+\text{b}^2)}{\text{a}^2+\text{b}^2}}$ $(\text{Using x}^2-\text{y}^2=(\text{x}-\text{y})(\text{x+y}))$ $=\sqrt{\frac{2\text{a}^2\times2\text{b}^2}{\text{a}^2+\text{b}^2}}$ $=\frac{2\text{ab}}{\text{a}^2+\text{b}^2}\cdots\text{(ii)}$ Now, $\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}$ $=\frac{\text{a}^2-\text{b}^2}{\frac{\text{a}^2+\text{b}^2}{\frac{2\text{ab}}{\text{a}^2+\text{b}^2}}}$ $=\frac{\text{a}^2-\text{b}^2}{2\text{}ab}$ $\sec=\text{x}\frac{1}{\cos\text{x}}=\frac{\text{a}^2+\text{b}^2}{2\text{ab}}$ (From(ii)) and $\text{cosec}\text{ x}=\frac{1}{\sin\text{x}}=\frac{\text{a}^2+\text{b}^2}{\text{a}^2-\text{b}^2}$ (From(i))
View full question & answer→Question 163 Marks
Prove that:
$\tan4\pi-\cos\Big(\frac{3\pi}{2}\Big)-\sin\Big(\frac{5\pi}{6}\Big)\cos\Big(\frac{2\pi}{3}\Big)=\frac{1}{4}$
Answer$\text{L.H.S}=\tan4\pi-\cos\Big(\frac{3\pi}{2}\Big)-\sin\Big(\frac{5\pi}{6}\Big)\cos\Big(\frac{2\pi}{3}\Big)$
$=\tan720^\circ-\cos270^\circ-\sin150^\circ\cos120^\circ$
$\tan4\pi-\cos\Big(\frac{3\pi}{2}\Big)-\sin\Big(\pi\frac{\pi}{6}\Big)\cos\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)$ $(\because\pi =180^\circ)$
$=0-0-\sin\frac{\pi}{6}\Big(-\sin\frac{\pi}{6}\Big)$ $\Big(\because\tan\text{n}\pi=0 \text{ for all n}\in \text{Z }\& \cos\frac{3\pi}{2}=0\Big)$
$=\sin^2\frac{\pi}{6}$
$=\Big(\frac{1}{2}\Big)^2$
$=\frac{1}{4}$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 173 Marks
Prove that:
$\tan225^\circ\cot405^\circ+\tan765^\circ\cot675^\circ=0$
Answer$\text{L.H.S}=\tan225^\circ\cot405^\circ+\tan765^\circ\cot675^\circ$
$=\tan\Big(\pi+\frac{\pi}{4}\Big)\cot\Big(2\pi+\frac{\pi}{4}\Big)+\tan\Big(4\pi+\frac{\pi}{4}\Big)\cot\Big(4\pi-\frac{\pi}{4}\Big)$
$=\tan\frac{\pi}{4}.\cot\frac{\pi}{4}+\tan\frac{\pi}{4}\times\Big(-\cot\frac{\pi}{4}\Big)$ $\Big(\because\cot\Big(4\pi-\frac{\pi}{4}\Big)=-\cot\frac{\pi}{4}\Big)$
$= 1.1 + 1. (-1)$
$= 0$
$\text{= R.H.S}$
$\text{Proved}$
View full question & answer→Question 183 Marks
Prove that:
$\cos570^\circ\sin510^\circ+\sin(-330^\circ)\cos(-390^\circ)=0$
Answer$\text{L.H.S}=\cos570^\circ\sin510^\circ+\sin(-330^\circ)\cos(-390^\circ)$
$\cos\Big(3\pi+\frac{\pi}{6}\Big)\sin\Big(3\pi-\frac{\pi}{6}\Big)-\sin330^\circ\cos390^\circ$$\begin{pmatrix}\because\sin(-\theta)=-\sin\theta\text{ and}\\\cos(-\theta)=\cos\theta\end{pmatrix}$
$=-\cos\frac{\pi}{6}\sin\frac{\pi}{6}-\sin\Big(2\pi-\frac{\pi}{6}\Big)\cos\Big(2\pi+\frac{\pi}{6}\Big)$
$=-\sin\frac{\pi}{6}\cos\frac{\pi}{6}+\sin\frac{\pi}{6}.\cos\frac{\pi}{6}$ $(\because\sin(2\pi-\theta)=-\sin\theta)$
$= 0$
$\text{= R.H.S}$
$\text{Proved}$
View full question & answer→Question 193 Marks
In a $\triangle\text{ABC},$ Prove that:
$\cos\Big(\frac{\text{A+B}}{2}\Big)=\sin\frac{\text{C}}{2}$
AnswerWe have $\text{A + B + C} = \pi$ $\big(\because$ sum of 3 angle of a triangle is $\pi=180^\circ\big)$
$\Rightarrow \text{A+B}={\pi}-\text{C}$
$$$\Rightarrow \frac{\text{A+B}}{2}=\frac{\pi-\text{C}}{2}$
$\Rightarrow \frac{\text{A+B}}{2}=\frac{\pi}{2}-\frac{\text{C}}{2}$
$\Rightarrow \cos=\Big(\frac{\text{A+B}}{2}\Big)=\cos\Big(\frac{\pi}{2}-\frac{C}{2}\Big)$
$\Rightarrow = \sin\frac{\text{C}}{2}$ $\Big(\because\cos\Big(\frac{\pi}{2}-\theta\Big)=\sin\theta\Big)$
Hence, $\cos\Big(\frac{\text{A+B}}{2}\Big)=\sin\frac{\text{C}}{2}$
$\text{Proved}$
View full question & answer→Question 203 Marks
Prove that:
$\tan\frac{11\pi}{3}-2\sin\frac{4\pi}{6}-\frac{3}{4}\text{cosec}^2\frac{\pi}{4}+4\cos^2\frac{17\pi}{6}=\frac{3-4\sqrt{3}}{2}$
Answer$\text{L.H.S}=\tan\frac{11\pi}{3}-2\sin\frac{4\pi}{6}-\frac{3}{4}\text{cosec}^2\frac{\pi}{4}+\cos^2\frac{17\pi}{6}$
$=\tan\Big(4\pi-\frac{\pi}{3}\Big)-2\sin\frac{2\pi}{3}-\frac{3}{4}\times(\sqrt{2})^2+4\cos^2\Big(3\pi-\frac{\pi}{6}\Big)$
$=-\tan\frac{\pi}{3}-2\sin\Big(\pi-\frac{\pi}{3}\Big)-\frac{3}{4}\times2+4\cos^2\frac{\pi}{6}$ $\Big(\because\tan\Big(4\pi-\frac{\pi}{3}=-\tan\frac{\pi}{3},\cos\Big(3\pi-\frac{\pi}{6}\Big)=-\cos\frac{\pi}{6}\Big)$
$=-\sqrt{3}=2\sin\frac{\pi}{3}-\frac{3}{2}+4\times\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=-\sqrt{3}-2\times\frac{\sqrt{3}}{2}-\frac{3}{2}+4\times\frac{3}{4}$
$=-\sqrt{3}-\sqrt{3}-\frac{3}{2}+3$
$=-2\sqrt{3}\frac{-3+6}{2}$
$=-2\sqrt{3}+\frac{3}{2}$
$=\frac{3-4\sqrt{3}}{2}$
$=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 213 Marks
Prove that:
$\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}=2$
Answer$\text{L.H.S}=\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}$
$=\sin^2\frac{\pi}{18}+\sin^2\frac{4\pi}{9}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}$
$=\sin^2\Big(\frac{\pi}{2}-\frac{4\pi}{9}\Big)+\sin^2\frac{4\pi}{9}+\sin^2\frac{\pi}{9}+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{9}\Big)$ $\big(\because\frac{\pi}{18}=\frac{\pi}{2}-\frac{4\pi}{9}\text{ and }\frac{7\pi}{18}=\frac{\pi}{2}-\frac{\pi}{9}\big)$
$=\cos^2\frac{4\pi}{9}+\sin^2\frac{4\pi}{9}+\sin^2\frac{\pi}{9}+\cos^2\frac{\pi}{9}$
$=1+1$ $(\because\sin^2\theta+\cos^2\theta=1)$
$= 2$
$= \text{R.H.S}$
$\text{proved}$
View full question & answer→Question 223 Marks
If $\cos\text{x}=-\frac{3}{5}$ and $\pi<\text{x}<\frac{3\pi}{2},$ Find the value of other of five trigonomentoric function and hence evaluate $\frac{\text{cosec x}+\cot \text{x}}{\sec\text{x}-\tan\text{x}}.$
AnswerWe have:$\cos\text{x} = -\frac{3}{5}$ and $\pi<\text{x}<\frac{3\pi}{2}$
Thus, x is in the third quadrant.
In the third quadrant, $\tan\text{x}$ and $\cot\text{x}$ are positive And, all the other four T- ratios are negative.
$\therefore \sin\text{x}= -\sqrt{1- \cos^2\text{x}} = -\sqrt{1-\Big(\frac{-3}{5}\Big)^2} = \frac{-4}{5}$
$\tan\text{x} =\frac{\sin\text{x}}{\cos\text{x}}= \frac{\frac{-4}{5}}{-\frac{3}{5}}=\frac{4}{3}$
$\cot\text{x}= \frac{1}{\tan\text{x}}= \frac {4}{3}=\frac{3}{4}$
$\sec\text{x}=\frac{1}{\cos\text{x}}=\frac{1}{\frac{-3}{5}}=\frac{-5}{3}$
$\text{cosec}\text{ x} =\frac{1}{\sin\text{x}}=\frac{1}{\frac{-4}{5}}=\frac{-5}{4}$
Now, $\frac{\text{cosec }\text{x} + \cot\text{x}}{\sec\text{x}-\tan\text{x}}=\frac{\frac{-5}{4}+\frac{3}{4}}{\frac{-5}{3}-\frac{4}{3}}$
$=\frac{\frac{-2}{4}}{\frac{-9}{3}}$
$=\frac{\frac{-1}{2}}{-3}=\frac{1}{6}$
View full question & answer→Question 233 Marks
If $\text{x}=\frac{\text{b}}{\text{a}},$ find the value of $\sqrt{\frac{\text{a+b}}{\text{a - b}}}+\sqrt{\frac{\text{a - b}}{\text{a+b}}}.$
AnswerGiven that: $\tan\text{x}=\frac{\text{b}}{\text{a}}$
$\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}$
$\frac{\text{a + b + a}-\text{b}}{\sqrt{(\text{a}-\text{b})(\text{a + b})}}=\frac{2\text{a}}{\sqrt{\text{a}^2-\text{b}^2}}=\frac{2\text{a}}{\text{a}\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}}$ $\Big[\because\tan\text{x}=\frac{\text{b}}{\text{a}}\Big]$
$=\frac{2}{1-\tan^2\text{x}}$
$=\frac{2}{\sqrt{1-\frac{\sin^2\text{x}}{\cos^2\text{x}}}}=\frac{2}{\frac{\sqrt{\cos^2\text{x}-\sin^2\text{x}}}{\cos\text{x}}}$
$=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$ $[\because\cos^2\text{x}-\sin^2\text{x}=\cos2\text{x}]$
Hence, $\sqrt{\frac{\text{a}+\text{b}}{\text{a}-\text{b}}}+\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}=\frac{2\cos\text{x}}{\sqrt{\cos2\text{x}}}$
View full question & answer→Question 243 Marks
In a $\triangle\text{ABC},$ Prove that:
$\tan\frac{\text{A+B}}{2}=\cot\frac{\text{C}}{2}$
AnswerWe have:
$\text{A + B + C} = \pi$ $\big(\because$ sum of 3 angle of a triangle is $\pi=180^\circ\big)$
$\Rightarrow\text{A+B}=\pi-\text{C}$
$\Rightarrow\frac{\text{A+B}}{2}=\frac{\pi-\text{C}}{2}$
$\Rightarrow\frac{\text{A+B}}{2}=\frac{\pi}{2}-\frac{\text{C}}{2}$
$\Rightarrow\tan\Big(\frac{\text{A+B}}{2}\Big)=\tan\Big(\frac{\pi}{2}-\frac{\text{C}}{2}\Big)$
$=\cot\frac{\text{C}}{2}$ $\Big(\because\tan\Big(\frac{\pi}{2}-\theta\Big)=\cot\theta\Big)$
Hence, $\tan\Big(\frac{\text{A+B}}{2}\Big)=\tan\Big(\frac{\pi}{2}-\frac{\text{C}}{2}\Big)$
$\text{Proved}$
View full question & answer→Question 253 Marks
Prove that: $\sin\frac{13\pi}{3}\sin\frac{2\pi}{3}+\cos\frac{4\pi}{3}\sin\frac{13\pi}{6}=\frac{1}{2}$
Answer$\text{L.H.S}=\sin\frac{13\pi}{3}\sin\frac{2\pi}{3}+\cos\frac{4\pi}{3}\sin\frac{13\pi}{6}$
$=\sin780^\circ\times\sin120^\circ+\sin240^\circ\sin390^\circ$
$=\sin\Big(4\pi+\frac{\pi}{3}\Big)\sin\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)+\cos\Big(\pi+\frac{\pi}{6}\Big)\sin\Big(2\pi+\frac{\pi}{6}\Big)$
$(\because\pi=180^\circ)$
$=\sin\frac{\pi}{3}\times\cos\frac{\pi}{6}-\cos\frac{\pi}{3}\times\Big(+\sin\frac{\pi}{6}\Big)$
$\left(\begin{array}{c}\because\sin\Big(4\pi+\frac{\pi}{3}\Big)=\sin\frac{\pi}{3}\\\&\sin\Big(3\pi-\frac{\pi}{3}\Big)=\sin\frac{\pi}{3}\end{array}\right)$
$=\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}-\frac{1}{2}\times\frac{1}{2}$
$=\frac{3}{4}-\frac{1}{4}$
$=\frac{1}{2}$
$=\text{R.H.S}$ Proved
View full question & answer→Question 263 Marks
Find the values of the other five trigonometric functions in the following:
$\cot\text{x}=-\frac{1}{2},$ x in quadrant II
AnswerWe have: $\cos\text{x}=-\frac{1}{2}$ and x are in the second quadrant, $\sin\text{x},$ and $\text{cosec}\text{ x}$ are positive and, $\tan\text{x},\cot\text{x},\cos\text{x}$ and $\sec\text{x}$ are negative.$\therefore\sin\text{x}=\sqrt{1-\cos\text{x}^2}=\sqrt{1-\Big(\frac{-1}{2}\Big)^2}=\frac{\sqrt3}{2}$
$\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}=\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}}=-\sqrt{3}$
$\cot\text{x}=\frac{1}{\tan\text{x}}=\frac{1}{-\sqrt3}=\frac{-1}{\sqrt3}$
$\sec\text{x}=\frac{1}{\cos\text{x}}=\frac{1}{\frac{-1}{2}}=-2$
$\text{cosec}\text{ x}=\frac{1}{\sin\text{x}}=\frac{1}{\frac{\sqrt3}{2}}=\frac{2}{\sqrt{3}}$
View full question & answer→Question 273 Marks
Prove that:
$\Big\{1+\cot\text{x}-\sec\Big(\frac{\pi}{2}+\text{x}\Big)\Big\}\Big\{1+\cot\text{x}+\sec\Big(\frac{\pi}{2}+\text{x}\Big)\Big\}=2\cot\text{x}$
Answer$\text{L.H.S}=\Big\{1+\cot\text{x}-\sec\Big(\frac{\pi}{2}+\text{x}\Big)\Big\}\Big\{1+\cot\text{x}+\sec\Big(\frac{\pi}{2}+\text{x}\Big)\Big\}$
$\{{1+\cot\text{x}-(-\text{cosec}\text{x})}\}\{1+\cot\text{x}-\text{cosec}\text{x}\}$ $\Big(\because\sec\Big(\frac{\pi}{2}+\text{x}\Big)=-\text{cosec }\text{x}\Big)$
$=\{(1+\cot)+\text{cosec}\}\{(1+\cot\text{x})-\text{cosec}\text{x}\}$
$=(1+\cot\text{x})^2-\text{cosec}^2\text{x}$
$=1+\cot^2\text{x}+2\cot\text{x}-\text{cosec}^2\text{x}$
$=\text{cosec}^2+2\cot\text{x}-\text{cosec}^2$ $(\because1+\cot^2\text{x}=\text{cosec}^2\text{x})$
$=2\cot\text{x}$
$\text{= R.H.S}$
$\text{Proved}$
View full question & answer→