Question 11 Mark
Prove the following : $\cos ^4 \theta-\sin ^4 \theta+1=2 \cos ^2 \theta$
Answer$\text { L.H.S. }=\cos ^4 \theta-\sin ^4 \theta+1$
$=\left(\cos ^2 \theta\right)^2-\left(\sin ^2 \theta\right)^2+1=\left(\cos ^2 \theta+\sin ^2 \theta\right) c\left(\cos ^2 \theta-\sin ^2 \theta\right)+1$
$=(1)\left(\cos ^2 \theta-\sin ^2 \theta\right)+1=\cos ^2 \theta+\left(1-\sin ^2 \theta\right)$
$=\cos ^2 \theta+\cos ^2 \theta=2 \cos ^2 \theta=\text { R.H.S. }$
View full question & answer→Question 21 Mark
Prove the following : $\sin ^4 \theta+\cos ^4 \theta=\sin ^4 \theta+\cos ^4 \theta$
Answer$\text { L.H.S. }=\sin ^4 \theta+\cos ^4 \theta$
$=\left(\sin ^2 \theta\right)^2+\left(\cos ^2 \theta\right)^2=\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta$
$\ldots\left[v a^2+b^2=(a+b)^2-2 a b\right]$
$=1-2 \sin ^2 \theta \cos ^2 \theta$
$=\text { R.H.S. }$
View full question & answer→Question 31 Mark
Prove the following : L.H.S. $=\left(\tan \theta+\frac{1}{\cos \theta}\right)^2+\left(\tan \theta-\frac{1}{\cos \theta}\right)^2=2\left(\frac{1+\sin ^2 \theta}{1-\sin ^2 \theta}\right)$
Answer$\text { L.H.S. }=\left(\tan \theta+\frac{1}{\cos \theta}\right)^2+\left(\tan \theta-\frac{1}{\cos \theta}\right)^2$
$=(\tan \theta+\sec \theta)^2+(\tan \theta-\sec \theta)^2$
$=\tan ^2 \theta+2 \tan \theta \sec \theta+\sec ^2 \theta$
$+\tan ^2 \theta-2 \tan \theta \sec \theta+\cdot \sec ^2 \theta$
$=2\left(\tan ^2 \theta+\sec ^2 \theta\right)$
View full question & answer→Question 41 Mark
Prove the following:
$\sin ^2 A \cos ^2 B+\cos ^2 A \sin ^2 B+\cos ^2 A \cos ^2 B+\sin ^2 A \sin ^2 B=1$
Answer$\text { L.H.S. }=\sin ^2 A \cos ^2 B+\cos ^2 A \sin ^2 B+\cos ^2 A \cos ^2 B+\sin ^2 A \sin ^2 B$
$=\sin ^2 A\left(\cos ^2 B+\sin ^2 B\right)+\cos ^2 A\left(\sin ^2 B+\cos ^2 B\right)$
$=\sin ^2 A(1)+\cos ^2 A(1)$
$=1=\text { R.H.S. }$
View full question & answer→Question 51 Mark
Show that $1 – 2sin θ cos θ ≥ 0$ for all $θ ∈ R.$
Answer$1 – 2 \sin \theta \cos \theta $
$= \sin^2 \theta + \cos^2 \theta – 2sin \theta \cos \theta $
$= (\sin \theta – \cos \theta )^2 \geq 0$ for all $θ ∈ R$
View full question & answer→Question 61 Mark
State the quadrant in which 6 lies if : sin θ > 0 and tan θ < 0
Answersin θ > 0
sin θ is positive in 1st and 2nd quadrants, tan θ < 0
tan θ is negative in 2nd and 4th quadrants.
∴ θ lies in the 2nd quadrant.
View full question & answer→Question 71 Mark
State the quadrant in which 6 lies if : sin θ < 0 and cos θ < 0
Answersin θ < 0
sin θ is negative in 3rd and 4th quadrants, cos θ < 0
cos θ is negative in 2nd and 3rd quadrants.
.’. θ lies in the 3rd quadrant.
View full question & answer→Question 81 Mark
State the quadrant in which 6 lies if : tan θ < 0 and sec θ > 0
Answertan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0
sec θ is positive in 1st and 4th quadrants.
∴ θ lies in the 4th quadrant.
View full question & answer→Question 91 Mark
State the signs of : sin 986°
Answer986° = 2x 360° + 266°
∴ 986° and 266° are co-terminal angles.
Since 180° < 266° < 270°,
266° lies in the 3rd quadrant.
∴ 986° lies in the 3rd quadrant.
∴ sin 986° is negative.
View full question & answer→Question 101 Mark
State the signs of : cot 1899°
Answer1899° = 5 x 360° + 99°
∴ 1899° and 99° are co-terminal angles.
Since 90° < 99° < 180°,
99° lies in the 2nd quadrant.
∴ 1899° lies in the 2nd quadrant.
∴ cot 1899° is negative.
View full question & answer→Question 111 Mark
State the signs of : cosec 520°
Answer520° =360° + 160°
∴ 520° and 160° are co-terminal angles.
Since 90° < 160° < 180°,
160° lies in the 2nd quadrant.
∴ 520° lies in the 2nd quadrant,
∴ cosec 520° is positive.
View full question & answer→Question 121 Mark
Prove the following : $\tan ^2 \theta-\sin ^2 \theta=\sin ^4 \theta \sec ^2 \theta$
Answer$ \text { L.H.S. }=\tan ^2 \theta-\sin ^2 \theta$
$=\frac{\sin ^2 \theta}{\cos ^2 \theta}-\sin ^2 \theta$
$=\sin ^2 \theta\left(\frac{1}{\cos ^2 \theta}-1\right)$
$=\frac{\sin ^2 \theta\left(1-\cos ^2 \theta\right)}{\cos ^2 \theta}$
$=\left(\sin ^2 \theta\right)\left(\sin ^2 \theta\right) \sec ^2 \theta$
$=\sin ^4 \theta \sec ^2 \theta$
$=\text { R.H.S } $
View full question & answer→Question 131 Mark
$\sin^4\theta + 2sin^2\theta \cos^2\theta = 1 – \cos^4\theta $
AnswerL.H.S. $= \sin^4\theta + 2sin^2\theta \cos^2\theta = \sin^2\theta (\sin^2\theta + 2cos^2\theta )$
$= (\sin^2\theta ) (\sin^2\theta + \cos^2\theta + \cos^2\theta ) = (1 – \cos^2\theta ) (1 + \cos^2\theta )$
$= 1 – \cos^4\theta $= R.H.S.
View full question & answer→Question 141 Mark
Eliminate θ from the following : $x=2+3 \cos \theta, \quad y=5+3 \sin \theta\
View full question & answer→Question 151 Mark
Eliminate θ from the following : $x= a \cos ^3 \theta, \quad y= b \sin ^3 \theta$
View full question & answer→Question 161 Mark
Eliminate θ from the following : x = a cosθ, y = b sinθ
Question 171 Mark
If $\tan A=\frac{4}{3}$, find the value of $\frac{2 \sin A-3 \cos A}{2 \sin A+3 \cos A}$
View full question & answer→Question 181 Mark
Prove that $\cos ^6 \theta+\sin ^6 \theta=1-3 \sin ^2 \theta \cos ^2 \theta$
View full question & answer→Question 191 Mark
Find the value of cos 765°
Question 201 Mark
Evaluate the following : $\sin \pi+2 \cos \pi+3 \sin \frac{3 \pi}{2}+4 \cos \frac{3 \pi}{2} \ -5 \sec \pi-6 \operatorname{cosec} \frac{3 \pi}{2}$
View full question & answer→Question 211 Mark
Evaluate the following : $\sin ^2 0+\sin ^2 \frac{\pi}{6}+\sin ^2 \frac{\pi}{3}+\sin ^2 \frac{\pi}{2}$
View full question & answer→Question 221 Mark
Evaluate the following : $4 \cos ^3 45^{\circ}-3 \cos 45^{\circ}+\sin 45^{\circ}$
View full question & answer→Question 231 Mark
Evaluate the following : cos30° × cos60° + sin30° × sin60°
Question 241 Mark
Find the value of $\sin \frac{41 \pi}{4}$.
View full question & answer→Question 251 Mark
For θ = 30°, Verify that sin2θ = 2sinθ cosθ
Question 261 Mark
Find the signs of the following : cot (−206°)
Question 271 Mark
Find the signs of the following : cos 400°
Question 281 Mark
Find the signs of the following : sin 300°
Question 291 Mark
Find the values of : $\cot \frac{25 \pi^e}{3}$
AnswerWe know that cotangent function is periodic with period $\pi$. $\cot \frac{25 \pi}{3}=\cot \left(8 \pi+\frac{\pi}{3}\right)=\cot \frac{\pi}{3}=\frac{1}{\sqrt{3}}$
View full question & answer→Question 301 Mark
Find the values of : $\cos 1140^{\circ}$
AnswerWe know that cosine function is periodic with period $2 \pi$.
$ \cos 1140^{\circ}=\cos \left(3 \times 360^{\circ}+60^{\circ}\right)$
$=\cos 60^{\circ}=\frac{1}{2} $
View full question & answer→Question 311 Mark
Find the values of : $\sin \frac{19 \pi^c}{3}$
AnswerWe know that sine function is periodic with period $2 \pi$.
$\sin \frac{19 \pi}{3}=\sin \left(6 \pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
View full question & answer→Question 321 Mark
Find the polar co-ordinates of points whose Cartesian co-ordinates are : $(-\sqrt{3}, 1)$
Answer$(x, y)=(-\sqrt{3}, 1)$
$\therefore r=\sqrt{x^2+y^2}=\sqrt{3+1}=\sqrt{4}=2$
$\tan \theta=\frac{y}{x}=\frac{1}{-\sqrt{3}}=-\tan \frac{\pi}{6} $
Since the given point lies in the $2$ nd quadrant,
$ \tan \theta=\tan \left(\pi-\frac{\pi}{6}\right) \ldots[\because \tan (\pi-x)=-\tan \mathrm{x}]$
$\therefore \tan \theta=\tan \left(\frac{5 \pi}{6}\right)$
$\therefore \theta=\frac{5 \pi}{6}=150^{\circ}$
$\therefore$ the required polar co-ordinates are $\left(2,150^{\circ}\right)$
View full question & answer→Question 331 Mark
Find the polar co-ordinates of points whose Cartesian co-ordinates are : $(-1,-1)$
Answer$(x, y)=(-1,-1)$
$ \therefore r=\sqrt{x^2+y^2}=\sqrt{1+1}=\sqrt{2}$
$\tan \theta=\frac{y}{x}=\frac{-1}{-1}=1$
$\therefore \tan \theta=\tan \frac{\pi}{4} $
Since the given point lies in the $3$ rd quadrant,
$ \tan \theta=\tan \left(\pi+\frac{\pi}{4}\right) \ldots[\because \tan (n+x)=\tan x]$
$\therefore \tan \theta=\tan \left(\frac{5 \pi}{4}\right)$
$\therefore \theta=\frac{5 \pi}{4}=225^{\circ}$
$\therefore$ the required polar co-ordinates are $\left(\sqrt{2}, 225^{\circ}\right)$.
View full question & answer→Question 341 Mark
Find the polar co-ordinates of points whose Cartesian co-ordinates are : $(1, \sqrt{3})$
Answer$(x, y)=(1, \sqrt{3})$
$ \therefore r=\sqrt{x^2+y^2}=\sqrt{1+3}=\sqrt{4}=2$
$\tan \theta=\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}$
Since the given point lies in the 1st quadrant,
$\theta=60^{\circ} \ldots\left[\because \tan 60^{\circ}=\sqrt{3}\right]$
$\therefore$ the required polar co-ordinates are $\left(2,60^{\circ}\right)$.
View full question & answer→Question 351 Mark
Find the polar co-ordinates of points whose Cartesian co-ordinates are : $(5,5)$
Answer$ \text { i. }(x, y)=(5,5)$
$\therefore r=\sqrt{x^2+y^2}=\sqrt{25+25}$
$=\sqrt{50}=5 \sqrt{2}$
$\tan \theta=\frac{y}{x}=\frac{5}{5}=1 $
Since the given point lies in the 1st quadrant,
$\theta=45^{\circ} \ldots\left[\because \tan 45^{\circ}=1\right]$
$\therefore$ the required polar co-ordinates are $\left(5 \sqrt{2}, 45^{\circ}\right)$.
View full question & answer→Question 361 Mark
Find the Cartesian co-ordinates of points whose polar co-ordinates are $: (1, 180^\circ)$
Answer$(r, \theta)=\left(1,180^{\circ}\right)$
Using $x=r \cos \theta$ and $y=r \sin \theta$, where $(x, y)$ are the required cartesian co-ordinates, we get
$x=1\left(\cos 180^{\circ}\right) \text { and } y=1\left(\sin 180^{\circ}\right)$
$\therefore x=-1 \text { and } y=0$
$\therefore$ the required cartesian co-ordinates are $(-1,0)$.
View full question & answer→Question 371 Mark
Find the Cartesian co-ordinates of points whose polar co-ordinates are $: (3, 90^\circ)$
Answer$(r, \theta)=\left(3,90^{\circ}\right)$
Using $x=r \cos \theta$ and $y=r \sin \theta$, where $(x, y)$ are the required cartesian co-ordinates, we get
$ x=3 \cos 90^{\circ} \text { and } y=3 \sin 90^{\circ}$
$\therefore x=3(0)=0 \text { and } y=3(1)=3$
$\therefore$ the required cartesian co-ordinates are $(0,3)$.
View full question & answer→Question 381 Mark
Eliminate 0 from the following $: x = 3 – 4 \tan \theta,3y = 5 + 3\sec \theta$
Answer$2 x=3-4 \tan \theta \text { and } 3 y=5+3 \sec \theta$
$\therefore 2 x-3=-4 \tan \theta \text { and } 3 y-5=3 \sec \theta$
$\therefore \tan \theta=\frac{3-2 x}{4} \text { and } \sec \theta=\frac{3 y-5}{3} \theta$
We know that, $\sec ^2 \theta-\tan ^2 \theta=1$
$\therefore\left(\frac{3 y-5}{3}\right)^2-\left(\frac{3-2 x}{4}\right)^2=1$
$\therefore\left(\frac{3 y-5}{3}\right)^2-\left(\frac{2 x-3}{4}\right)^2=1 $
View full question & answer→Question 391 Mark
Eliminate 0 from the following $: x = 5 + 6 \operatorname{cosec} \theta,y = 3 + 8 \cot \theta$
Answer$ x=5+6 \operatorname{cosec} \theta \text { and } y=3+8 \cot \theta$
$\therefore x-5=6 \operatorname{cosec} \theta \text { and } y-3=8 \cot \theta$
$\therefore \operatorname{cosec} \theta=\frac{x-5}{6} \text { and } \cot \theta=\frac{y-3}{8} $
We know that,
$\operatorname{cosec}^2 \theta-\cot ^2 \theta=1$
$\therefore\left(\frac{x-5}{6}\right)^2-\left(\frac{y-3}{8}\right)^2=1$
View full question & answer→Question 401 Mark
Eliminate 0 from the following : x = 4cos θ – 5sin θ, y = 4sin θ + 5cos θ
Answer$x=4 \cos \theta-5 \sin \theta \ldots$ (i)
$y=4 \sin \theta+5 \cos \theta$... (ii)
Squaring (i) and (ii) and adding, we get
$ x^2+y^2=(4 \cos \theta-5 \sin \theta)^2+(4 \sin \theta+5 \cos \theta)^2$
$=16 \cos ^2 \theta-40 \sin \theta \cos \theta+25 \sin ^2 \theta+16 \sin ^2 \theta+40 \sin \theta \cos \theta+25 \cos ^2 \theta$
$=16\left(\sin ^2 \theta+\cos ^2 \theta\right)+25\left(\sin ^2 \theta+\cos ^2 \theta\right)$
$=16(1)+25(1)$
$=41 $
View full question & answer→Question 411 Mark
Eliminate 0 from the following : x = 6cosec θ,y = 8cot θ
Answer$x=6 \operatorname{cosec} \theta$ and $y=8 \cot \theta$
$\therefore \operatorname{cosec} \theta=$ and $\cot \theta=$
We know that,
$\operatorname{cosec}^2 \theta-\cot ^2 \theta=$
$\therefore \quad\left(\frac{x}{6}\right)^2-\left(\frac{y}{8}\right)^2=1$
$\therefore \quad \frac{x^2}{36}-\frac{y^2}{64}=1$
$16 x^2-9 y^2=576$
View full question & answer→Question 421 Mark
Eliminate 0 from the following $: x = 3\sec \theta, y = 4 \tan \theta$
Answer$x=3 \sec \theta, y=4 \tan \theta$
$\therefore \sec \theta=\frac{x}{3}$ and $\tan \theta=\frac{y}{4}$
We know that,
$\sec ^2 \theta-\tan ^2 \theta=1$
$\therefore \quad\left(\frac{x}{3}\right)^2-\left(\frac{y}{4}\right)^2=1$
$\therefore \quad \frac{x^2}{9}-\frac{y^2}{16}=1$
$\therefore 16 x^2-9 y^2=144 $
View full question & answer→Question 431 Mark
If $\tan \theta=\frac{1}{2}$, evaluate $\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}$
AnswerGiven, $\tan \theta=\frac{1}{2}$
$\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}=\frac{\frac{2 \sin \theta}{\cos \theta}+3}{4+\frac{3 \sin \theta}{\cos \theta}}$
..[Dividing numerator anddenominator by $\cos \theta$ ]
$=\frac{2 \tan \theta+3}{4+3 \tan \theta}$
$=\frac{2\left(\frac{1}{2}\right)+3}{4+3\left(\frac{1}{2}\right)}=\frac{4}{\left(\frac{11}{2}\right)}=\frac{8}{11} $
View full question & answer→Question 441 Mark
Using tables evaluate the following : $\cos ^2 0+\cos ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}+\cos ^2 \frac{\pi}{2}$
AnswerWe know that,
$\cos 0=1, \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \cos \frac{\pi}{3}=\frac{1}{2}$
$\cos \frac{\pi}{2}=0 \quad$
$\cos ^2 0+\cos ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}+\cos ^2 \frac{\pi}{2}$
$=1+\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2+0$
$=1+\frac{3}{4}+\frac{1}{4}=2 $
View full question & answer→Question 451 Mark
Using tables evaluate the following : $4 \cot 45^{\circ}-\sec 260^{\circ}+\sin 30^{\circ}$
AnswerWe know that,
$\cot 45^{\circ}=1, \sec 60^{\circ}=2, \sin 30^{\circ}=1 / 2$
$4 \cot 45^{\circ}-\sec ^2 60^{\circ}+\sin 30^{\circ}$
$=4(1)-(2)^2+\frac{1}{2}$
$=4-4+\frac{1}{2}=\frac{1}{2} $
View full question & answer→Question 461 Mark
Evaluate each of the following : sin 30° x cos 45° x lies tan 360°
AnswerWe know that,
$ \sin 30^{\circ}=\frac{1}{2}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}=, \tan 360^{\circ}=0$
$\sin 30^{\circ} \times \cos 45^{\circ} \times \tan 360^{\circ}$
$=\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)=0 $
View full question & answer→Question 471 Mark
Evaluate each of the following : cosec 45° + cot 45° + tan 0°
AnswerWe know that,
$ \operatorname{cosec} 45^{\circ}=\sqrt{2}, \cot 45^{\circ}=1, \tan 0^{\circ}=0$
$\operatorname{cosec} 45^{\circ}+\cot 45^{\circ}+\tan 0^{\circ}$
$=\sqrt{2}+1+0=\sqrt{2}+1 $
View full question & answer→Question 481 Mark
Evaluate each of the following : sin 30° + cos 45° + tan 180°
AnswerWe know that,
$ \sin 30^{\circ}=1 / 2, \cos 45^{\circ}=\frac{1}{\sqrt{2}}=, \tan 180^{\circ}=0$
$\sin 30^{\circ}+\cos 45^{\circ}+\tan 180^{\circ}$
$=\frac{1}{2}+\frac{1}{\sqrt{2}}+0=\frac{\sqrt{2}+1}{2} $
View full question & answer→Question 491 Mark
State the quadrant in which 6 lies if : cos θ < 0 and tan θ > 0
Answercos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.
View full question & answer→Question 501 Mark
State the quadrant in which 6 lies if : sin θ < 0 and tan θ > 0
Answersin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.
View full question & answer→