# Class 11 Maths Ncert Solutions Ex 3.2

## Class 11 Maths Ncert Solutions Chapter 3 Ex 3.2

Q.1: Calculate the values of five trigonometric func. if cosy$\cos y$ = 12$-\frac{1}{2}$ and y lies in 3rd quadrant.

Sol:

(i)  sec y :

Since, cos y = 12$\frac{1}{2}$

Therefore, sec y = 1cosy$\frac{1}{\cos y}$ = 1(12)$\frac{1}{\left (-\frac{1}{2} \right )}$

Hence, sec y = -2

(ii)  sin y :

Since, sin2y+cos2y=1$\sin ^{2}y \;+ \;\cos ^{2}y \;= \;1$

Therefore, sin2y=1cos2y$\sin ^{2}y \;= \;1 \;- \;\cos ^{2}y$

$\Rightarrow$ sin2y=1(12)2$\sin ^{2}y \;= \;1 \;- \;\left (-\frac{1}{2} \right )^{2}$

$\Rightarrow$ sin2y=114$\sin ^{2}y \;= \;1 \;- \;\frac{1}{4}$

$\Rightarrow$ sin2y=34$\sin ^{2}y \;= \;\frac{3}{4}$

$\Rightarrow$ siny=±32$\sin y \;= \;\pm \frac{\sqrt{3}}{2}$

Since, y lies in the third quadrant, the value of sin y will be negative.

Therefore, sin y = 32$\frac{\sqrt{3}}{2}$

(iii)  cosec y = 1siny$\frac{1}{\sin y}$ = 1(32)$\frac{1}{\left (-\frac{\sqrt{3}}{2} \right )}$ = 23$-\frac{2}{\sqrt{3}}$

Therefore, cosec y = 23$-\frac{2}{\sqrt{3}}$

(iv)  tan y = sinycosy$\frac{\sin y}{\cos y}$ = tany=(32)(12)$\tan y=\frac{\left (-\frac{\sqrt{3}}{2} \right )}{\left (-\frac{1}{2} \right )}$ = 3$\sqrt{3}$

Therefore, tan y = 3$\sqrt{3}$

(v)  cot y = 1tany$\frac{1}{\tan y}$ = 13$\frac{1}{\sqrt{3}}$

Therefore, cot y = 13$\frac{1}{\sqrt{3}}$

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = 35$\frac{3}{5}$, where y lies in second quadrant.

Sol:

sin y = 35$\frac{3}{5}$

Therefore, cosec y = 1Siny$\frac{1}{Sin\; y}$ = 135=53$\frac{1}{\frac{3}{5}} = \frac{5}{3}$

Since, sin2y+cos2y=1$sin^{2}\;y \; + \; cos^{2}\;y = 1$

$\Rightarrow$  cos2y=1sin2y$cos^{2} y = 1 – sin^{2} y$

$\Rightarrow$  cos2y=1(35)2$cos^{2} y = 1 – \left ( \frac{3}{5} \right )^{2}$

$\Rightarrow$ cos2y=1925$cos^{2} y = 1 – \frac{9}{25}$

$\Rightarrow$ cos2y=1625$cos^{2} y = \frac{16}{25}$

$\Rightarrow$ cos y = ±45$\pm \frac{4}{5}$

Since, y lies in the 2nd quadrant, the value of cos y will be negative,

Therefore, cos y = – 45$\frac{4}{5}$

$\Rightarrow$ sec y = 1cosy=1(45)=54$\frac{1}{cos \; y} = \frac{1}{\left (- \frac{4}{5} \right )} =\; – \frac{5}{4}$

$\Rightarrow$ tan y = sinycosy=(35)(45)=34$\frac{sin \; y}{cos \; y} = \frac{\left (\frac{3}{5} \right ) }{\left (-\frac{4}{5} \right ) } = \;- \frac{3}{4}$

$\Rightarrow$ cot y = 1tany=43$\frac{1}{tan \; y} =\;- \frac{4}{3}$

Q.3: Find the values of other five trigonometric functions if coty=34$cot \; y = \frac{3}{4}$, where y lies in the third quadrant.

Sol:

cot y = 34$\frac{3}{4}$

Since, tan y = 1coty=134=43$\frac{1}{cot \; y}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

Since, 1+tan2y=sec2y$1 + tan^{2}y = sec^{2}y$

$\Rightarrow$  1+(43)2=sec2y$1 + \left ( \frac{4}{3} \right )^{2} = sec^{2}y$

$\Rightarrow$  1+169=sec2y$1 + \frac{16}{9} = sec^{2}y$

$\Rightarrow$  259=sec2y$\frac{25}{9} = sec^{2}y$

$\Rightarrow$  sec y = ±53$\pm \frac{5}{3}$

Since, y lies in the 3rd quadrant, the value of sec y will be negative.

Therefore, sec y = 53$\;- \frac{5}{3}$

cos y = 1secy=1(53)=35$\frac{1}{sec \; y} = \frac{1}{\left (- \frac{5}{3} \right )} = – \frac{3}{5}$

Since, tan y = sinycosy$\frac{sin \; y}{cos \; y}$

$\Rightarrow$  43=siny(35)$\frac{4}{3} = \frac{sin \; y}{\left (- \frac{3}{5} \right )}$

$\Rightarrow$ sin y = (43)×(35)=45$\left (\frac{4}{3} \right ) \times \left (- \frac{3}{5} \right ) = -\frac{4}{5}$

$\Rightarrow$  cosec y = 1siny=54$\frac{1}{sin \; y} = -\frac{5}{4}$

Q.4: Find the values of other five trigonometric if secy=135$sec \; y = \frac{13}{5}$, where y lies in the fourth quadrant.

Sol:

sec y = 135$\frac{13}{5}$

cos y = 1secy=1(135)=513$\frac{1}{sec \; y} = \frac{1}{\left (\frac{13}{5}\right )} = \frac{5}{13}$

Since, sin2y+cos2y=1$sin^{2}y + cos^{2}y = 1$

$\Rightarrow$   sin2y=1cos2y$sin^{2}y = 1 – cos^{2}y$

$\Rightarrow$   sin2y=1(513)2$sin^{2}y = 1 – \left ( \frac{5}{13} \right )^{2}$

$\Rightarrow$   sin2y=125169=144169$sin^{2}y = 1 – \frac{25}{169} = \frac{144}{169}$

$\Rightarrow$ sin y = ±1213$\pm \frac{12}{13}$

Since, y lies in the 4th quadrant, the value of sin y will be negative.

Therefore, sin y = 1213$– \frac{12}{13}$

$\Rightarrow$ cosec y = 1siny=1(1213)=1312$\frac{1}{sin \; y} = \frac{1}{\left (- \frac{12}{13} \right )} = \;- \frac{13}{12}$

$\Rightarrow$ tan y = sinycosy=(1213)(513)=125$\frac{sin \; y}{cos \; y} = \frac{\left (- \frac{12}{13} \right )}{\left (\frac{5}{13} \right )} =\; -\frac{12}{5}$

$\Rightarrow$ cot y = 1tany=1(125)=512$\frac{1}{tan \; y} = \frac{1}{\left ( -\frac{12}{5} \right )} = \;-\frac{5}{12}$

Q.5: Find the values of other five trigonometric function if tan y = 512$– \frac{5}{12}$ and y lies in second quadrant.

Sol:

tan  y = 512$-\frac{5}{12}$   [Given]

And, cot y = 1tany=1(512)=125$\frac{1}{tan \; y} = \frac{1}{ \left (- \frac{5}{12} \right )} = – \frac{12}{5}$

Since, 1+tan2y=sec2y$1 + tan^{2} y = sec^{2} y$

Therefore, 1+(512)2=sec2y$1 + \left ( – \frac{5}{12} \right )^{2} = sec^{2} y$

$\Rightarrow$   sec2y=1+(25144)$sec^{2} y = 1 + \left ( \frac{25}{144} \right )$

$\Rightarrow$  sec2y=169144$sec^{2} y = \frac{169}{144}$

$\Rightarrow$ sec y = ±1312$\pm \frac{13}{12}$

Since, y lies in the 2nd quadrant, the value of sec y will be negative.

Therefore, sec  y = 1312$– \frac{13}{12}$

cos y = 1secy=1(1312)=(1213)$\frac{1}{sec \; y } = \frac{1}{\left (- \frac{13}{12} \right )} = \left (- \frac{12}{13} \right )\\$

Since, tany=sinycosy$tan \; y = \frac{sin \; y}{cos \; y}$

$\Rightarrow$  512=siny(1213)$-\frac{5}{12} = \frac{sin \; y}{\left (- \frac{12}{13} \right )}\\$

$\Rightarrow$ sin y = (512)×(1213)=513$\left (-\frac{5}{12} \right ) \times \left (- \frac{12}{13}\right ) = \frac{5}{13}\\$

$\Rightarrow$ cosec y = 1siny=1(513)=135$\frac{1}{sin \; y} = \frac{1}{\left ( \frac{5}{13} \right )} = \frac{13}{5}$

Q.6: Calculate the value of trigonometric function sin 765°.

Sol:

The values of sin y repeat after an interval of 360° or 2n.

Therefore, sin 765° = sin ( 2 × 360° + 45° ) = sin 45° = 12$\frac{1}{\sqrt{2}}$

Q.7: Calculate the value of trigonometric function cosec [-1410°]

Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, cosec [-1410°] = cosec [-1410° + 4 × 360°] = cosec [ -1410° + 1440°] = cosec 30° = 2

Q.8: Calculate the value of the trigonometric function tan19π3$\tan \frac{19\pi }{3}$.

Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, tan19π3$\tan \frac{19\pi }{3}$ = tan613π$\tan 6\frac{1}{3}\pi$ = tan(6π+π3)$\tan \left ( 6\pi + \frac{\pi }{3} \right )$ = tanπ3$\tan \frac{\pi }{3}$ = tan 60° = 3$\sqrt{3}$

Q.9: Calculate the value of the trigonometric function sin(11π3)$\sin \left ( -\frac{11\pi }{3} \right )$.

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, sin(11π3)$\sin \left ( -\frac{11\pi }{3} \right )$ = sin(11π3+2×2π)$\sin \left ( -\frac{11\pi }{3} + 2 \times 2\pi \right )$ = sinπ3$\sin \frac{\pi }{3}$ = 32$\frac{\sqrt{3}}{2}$

Q.10: Calculate the value of the trigonometric function cot(15π4)$\cot \left ( -\frac{15\pi }{4} \right )$.

Sol:

It is known that the values of tany$\tan y$ repeat after an interval of 180$180^{\circ}$ or n.

Therefore, cot(15π4)$\cot \left ( -\frac{15\pi }{4} \right )$ = cot(15π4+4π)$\cot \left ( -\frac{15\pi }{4} + 4\pi \right )$ = cotπ4$\cot \frac{\pi }{4}$ = 1