# RS Aggarwal Class 10 Solutions Trigonometric Ratios

## RS Aggarwal Class 10 Solutions Chapter 5

Q1:

If $\sin \theta =\frac{\sqrt{3}}{2}$, find the value of all T-ratios of $\theta$.

Sol:

Given:

$\sin \theta =\frac{\sqrt{3}}{2}$

Let us draw a $\bigtriangleup ABC$ in which $\angle B=90^{\circ}$ and $\angle B= 90^{\circ}=\theta$.

Then, $\sin \theta =\frac{BC}{AC}=\frac{\sqrt{3}}{2}$

Let BC =$\sqrt{3}k$

And AC= 2k

where k is positive

By pythagoras theorem, we have

$AC^{2}=AB^{2}+BC^{2}$
$\Rightarrow AB^{2}=AC^{2}-BC^{2}$
$\Rightarrow AB^{2}=(2k)^{2}-(\sqrt{3}k)^{2}$
$\Rightarrow AB^{2}=(4k^{2}-3k^{2})$
$\Rightarrow AB= \sqrt{k^{2}}=k$

Therefore, $\sin \theta = \frac{BC}{AC} = \frac{\sqrt{3}k}{2k} =\frac{\sqrt{3}}{2}$

$\cos \theta =\frac{AB}{AC}=\frac{k}{2k}=\frac{1}{2}$
$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\sqrt{3}}{2}\times \frac{2}{1}=\sqrt{3},$
$cosec \theta =\frac{1}{\sin \theta }=\frac{2}{\sqrt{3}},$

$\sec \theta =\frac{1}{\cos \theta }=\frac{2}{1}= 2$ and

$\cot \theta =\frac{1}{\tan \theta }=\frac{1}{\sqrt{3}}$

Q2:

If $\cos \theta =\frac{7}{25}$, find the value of all T-ratios of $\theta$.

Sol:

$\cos \theta =\frac{\sqrt{7}}{25}$

Let us draw a $\bigtriangleup ABC$ in which $\angle B = 90^{\circ}$ and $\angle BAC= \theta$

Let AB= 7k and AC= 25k,

Where k is positive

By pythagoras theorem, we have

$AC^{2}=AB^{2}+BC^{2}$
$\Rightarrow BC^{2}=AC^{2}-AB^{2}$
$\Rightarrow BC^{2}=(25k)^{2}-(7k)^{2}$
$=(625k^{2})-(49k^{2})$

= $=576k^{2}$

$\Rightarrow BC= \sqrt{576k^{2}}=24k$

Therefore, $\sin \theta = \frac{BC}{AC} = \frac{24k}{25k} = \frac{24}{25},$

$\cos \theta =\frac{7}{5}$ ??????( given )

$\tan \theta =\frac{\sin \theta }{\cos \theta }= \frac{24}{25}\times \frac{25}{7}=\frac{24}{7}$
$cosec \theta =\frac{1}{\sin \theta }=\frac{25}{24}$

$\sec \theta =\frac{1}{\cos \theta }=\frac{25}{7}$ and

$\cot \theta =\frac{1}{\tan \theta }=\frac{7}{24}$

Q3:

If $\tan \theta =\frac{15}{8}$, find the value of all T-ratios of $\theta$.

Sol:

Given:

$\tan \theta =\frac{BC}{AB} = \frac{15}{8}$

Let AB=15k and AC=8k,

Where k is positive

Let us draw a $\bigtriangleup ABC$ in which $\angle B=90^{\circ}$ and $\angle BAC=90^{\circ} \; and \; \angle BAC=\theta$

By pythagoras theorem, we have

$AC^{2}=AB^{2}+BC^{2}$
$=(8k)^{2}+(15)k^{2}$
$=(64k^{2})+(225k^{2})$

= $=289k^{2}$

AC=17k

Therefore, $\sin \theta= \frac{BC}{AC} = \frac{15k} {17k} = \frac{15} {17}$

$\cos \theta= \frac{AB} {AC} = \frac{8k} {17k} = \frac{8} {17}$

$\tan \theta= \frac{15} {8}$ ????(given)

$cosec \theta = \frac{1}{\sin \theta } = \frac{17}{15}$
$\sec \theta= \frac{1}{\cos \theta } = \frac{17}{8}$

and $\cot \theta= \frac{1}{\tan \theta } = \frac{8}{15}$

Q4:

If $\cot \theta = 2$, find the value of all T-ratios of $\theta$.

Sol:

Given:

$\cot \theta =\frac{AB}{BC}=\frac{2}{1}$

Let AB = 2k and AC = 1k,

Where k is positive

Let us draw a $\bigtriangleup ABC$ in which $\angle B=90^{\circ}$ and $\angle BAC = \theta$

By pythagoras theorem, we have

$AC^{2} = AB^{2}+BC^{2}$
$AC^{2} = (2k)^{2}+(k)^{2}$
$AC^{2} = 4k^{2}+k^{2}=5k^{2}$

Therefore, $AC = \sqrt{5k^{2}} = \sqrt{5}k$

$\sin \theta = \frac{BC}{AC} = \frac{k}{\sqrt{5}k} = \frac{1}{\sqrt{5}}$
$\cos \theta = \frac{AB}{AC} = \frac{2k}{\sqrt{5}k} = \frac{2} {\sqrt{5}}$
$\tan \theta = \frac{1}{\cot \theta } = \frac{1}{2}$
$\cot \theta = 2$
$cosec \theta = \frac{1}{\sin \theta } = \sqrt{5}$
$sec \theta = \frac{1}{\cos \theta } = \frac{\sqrt{5}} {2}$

Q5:

If $cosec \theta = ??\sqrt{10}$, find the value of all T-ratios of $\theta$.

Sol:

Given:

$cosec \theta =\frac{AC}{BC}=\frac{\sqrt{10}}{1}$

Let $AB = \sqrt{10}k$ and AC = 1k,

Where k is positive

Let us draw a $\bigtriangleup ABC$ in which $\angle B=90^{\circ}$ and $\angle BAC = \theta$

By pythagoras theorem, we have

$AC^{2} = AB^{2}+BC^{2}$
$AB^{2} = AC^{2} – BC^{2}$
$AB^{2} = (\sqrt{10}k)^{2} – (k)^{2}$
$AB^{2} = 10k^{2} – k^{2} = 9k^{2}$

Therefore, $AB = \sqrt{9k^{2}} = 3k$

$\sin \theta = \frac{BC}{AC} = \frac{1}{\sqrt{10}}$
$\cos \theta = \frac{AB}{AC} = \frac{3k}{\sqrt{10}k} = \frac{3} {\sqrt{10}}$

$cosec \theta = \sqrt{10}$ ????????????????(given)

$sec \theta = \frac{1}{\cos \theta } = \frac{\sqrt{10}} {3}$
$\tan \theta = \frac{\sin \theta }{\cos \theta }= \left ( \frac{1}{\sqrt{10}} \times \frac{\sqrt{10}}{3} \right ) = \frac{1}{3}$
$\cot \theta = \frac{1 }{\tan \theta }= 3$

Q6:

If $\sin \theta = \frac{a^{2} – b^{2}}{a^{2} + b^{2}}$, find the values of all T-ratio of $\theta$.

Sol:

Given:

We have,

$\sin \theta = \frac{a^{2} – b^{2}}{a^{2} + b^{2}}$

As,

$\cos ^{2}\theta =1-\sin ^{2}\theta$
$= 1- \left ( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \right )^{2}$
$= \frac {(a^{2}+b^{2})^{2} – (a^{2} – b^{2})^{2}}{{(a^{2}+b^{2}}) ^{2}}$
$= \frac {[(a^{2}+b^{2}) – (a^{2} – b^{2})][(a^{2}+b^{2}) + (a^{2} – b^{2})]}{{(a^{2}+b^{2}}) ^{2}}$
$= \frac {[a^{2}+b^{2} – a^{2} + b^{2}][a^{2}+b^{2} + a^{2} – b^{2}]}{{(a^{2}+b^{2}}) ^{2}}$
$= \frac {[2b^{2}][2a^{2}]}{{(a^{2}+b^{2}}) ^{2}}$
$\cos ^{2} \theta = \frac {4a^{2}b^{2}}{{(a^{2}+b^{2}}) ^{2}}$
$\cos \theta = \sqrt{\frac {4a^{2}b^{2}}{{(a^{2}+b^{2}}) ^{2}}}$
$\cos \theta = \frac {2ab}{{(a^{2}+b^{2}}) }$

Also,

$\tan \theta = \frac {\sin \theta}{\cos \theta }$
$\tan \theta = \frac {\frac{a^{2}-b^{2}}{a^{2}+b^{2}}}{\frac{2ab}{a^{2}+b^{2}}}$
$= \frac{a^{2} – b^{2}}{2ab}$
$cosec\; \theta = \frac{1}{\sin \theta}$
$= \frac{1}{\frac{a^{2} – b^{2}}{a^{2} + b^{2}}}$
$= \frac{a^{2} + b^{2}}{a^{2} – b^{2}}$

Also,

$\sec \theta = \frac{1}{\cos \theta}$
$= \frac{1}{\frac {2ab}{{(a^{2}+b^{2}}) }}$
$= \frac{(a^{2}+b^{2})}{2ab}$

And

$\cot \theta =\frac{1}{\tan \theta }$
$= \frac{1}{\frac{a^{2}-b^{2}}{2ab}}$
$= \frac{2ab}{a^{2}-b^{2}}$

Q7:

If $15\cot \theta = 8$, find the value of $\sin \theta \; and \; \sec \theta$.

Sol:

Given:

$\cot \theta =\frac{AB}{BC}=\frac{15}{8}$

Let AB = 15k and BC = 8k,

Where k is positive

Let us draw a $\bigtriangleup ABC$ in which $\angle B=90^{\circ}$ and $\angle BAC = \theta$

By pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2}$
$AC^{2} = (15k)^{2} + (8k)^{2}$
$AC^{2} = (225 k)^{2} + 64k^{2}$
$AC^{2} = 289k^{2} = 17k^{2}$

Therefore, $AC = \sqrt{289k^{2}} = 17k$

$\sin \theta = \frac{BC}{AC} = \frac{8k}{17k} \frac{8} {17}$
$\cos \theta = \frac{AB}{AC} = \frac{15k}{17k} = \frac{15} {17}$

Therefore, $\frac{(2 + 2sin\theta)(1- \sin \theta)}{(1+ \cos \theta) (2-2 \cos \theta)} = \frac{(2+2\times \frac{8}{17}) (1-\frac{8}{17})} {(1+\frac{15}{17}) (2-2\times \frac{15} {17})}$

$\frac{\frac{50}{17}\times \frac{9}{17}}{\frac{32}{17}\times \frac{4}{17}}=\frac{50\times 9}{32\times 4}=\frac{225}{64}$

Q8:

If sin A = $\frac{9}{41}$, find the values of cos A and tan A.

Sol:

We have sin A = $\frac{9}{41}$

As,

$cos^{2}A=1-sin^{2}A$
$=1-\left ( \frac{9}{41} \right )^{2}$
$=1-\frac{81}{1681}$
$=\frac{1681-81}{1681}$
$\Rightarrow cos^{2}A=\frac{1600}{1681}$
$\Rightarrow cosA=\sqrt{\frac{1600}{1681}}$
$\Rightarrow cosA=\frac{40}{41}$

Also.

$tanA=\frac{sinA}{cosA}$

= $\frac{\left ( \frac{9}{41} \right )}{\left ( \frac{40}{41} \right )}$

$=\frac{9}{40}$

Q9:

If $\cos \theta = 0.6$, show that $(5\sin \theta -3\tan \theta) = 0$.

Sol:

Given:

$\cos \theta = 0.6 = \frac{6} {10} = ??\frac{3} {5} ??$

Let us draw a $\bigtriangleup ABC$ in which $\angle B =90^{\circ} \; and \; \angle BCA = \theta$

Then, $\cos \theta = \frac{AB}{AC}= \frac{3}{5}$

Let AB = 3k

and AC = 5k

Where k is positive.

By pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2}$
$BC^{2}= AC^{2} – AB^{2}$
$=\left [ (5k)^{2} – (3k)^{2} \right ] = 16k^{2}$
$BC^{2} = 16k^{2}$
$BC= 4k$
$\sin \theta= \frac{AB}{AC} = \frac{4k} {5k} = \frac{4} {5}$
$\cos \theta =\frac{3} {5}$
$\tan \theta = \frac{\sin \theta }{\cos \theta }= \frac{4}{5}\times \frac{5}{3}= \frac{4}{3}$
$\Rightarrow (5\sin \theta -3\tan \theta) = 5\times \frac{4}{5} – 3\times \frac{4}{3} = 0$

Hence, $(5\sin \theta -3\tan \theta) = 0$

Q10:

If $cosec \theta = 2$, show that $\cot \theta +\frac{\sin \theta }{1+\cos \theta } = 2$

Sol:

Given:

$cosec \theta =\frac{AC}{BC}=\frac{2}{1}$

Let $AC = 2k$ and BC = 1k,

Where k is positive

Let us draw a $\bigtriangleup ABC$ in which $\angle B=90^{\circ}$ and $\angle BAC = \theta$

By pythagoras theorem, we have

$AC^{2} = AB^{2}+BC^{2}$
$AB^{2} = AC^{2} – BC^{2}$
$AB^{2} = (2k)^{2} – (k)^{2}$
$AB^{2} = 4k^{2} – k^{2} = 3k^{2}$

Therefore, $AB = \sqrt{3k^{2}} = ??\sqrt{3} k$

$\sin \theta = \frac{BC}{AC} = \frac{1}{2}$
$\cos \theta = \frac{AB}{AC} = \frac{ \sqrt{3}k} {2k} = \frac{\sqrt{3}} {2}$
$\cot \theta = \frac{\sin \theta } {\cos \theta } = \frac{\sqrt{3}} {2} \times \frac{2}{1} = \sqrt{3}$
$\Rightarrow \left [ \cot \theta +\frac{\sin \theta }{1+\cos \theta } \right ]= \left [ \sqrt{3} + \frac{\frac{1}{2}}{1+\frac{3}{2}} \right ]$
$= \left [ \sqrt{3}+\frac{1}{2+\sqrt{3}} \right ]= \left [ \frac{2\sqrt{3}+3+1}{2+\sqrt{3}} \right ]$
$= \left [ \frac{2\sqrt{3}+4}{2+\sqrt{3}} \right ] = 2\left [ \frac{\sqrt{3}+2}{2+\sqrt{3}} \right ] = 2$

Hence, $\cot \theta +\frac{\sin \theta }{1+\cos \theta } =2$

Q11:

If $\tan \theta = \frac{1}{\sqrt{7}}$, $\frac{(cosec^{2}\theta – \sec ^{2} \theta )}{(cosec^{2}\theta + \sec ^{2} \theta )} = \frac{3}{4}$ .

Sol:

Given:

$\tan \theta =\frac{BC}{AB}= \frac{1} {\sqrt{7}}$

Let BC = 1k and AB = $\sqrt{7}k$,

Where k is positive

Let us draw a $\bigtriangleup ABC$ in which $\angle B=90^{\circ}\; and \; \angle BAC=\theta$

By pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2}$
$= (\sqrt{7}k)^{2}+(k)^{2}$
$= (7k^{2})+(k^{2})$
$= 8k^{2}$

AC = $2\sqrt{2}k$

Therefore, $cosec \theta= \frac{AC}{BC} = \frac{2\sqrt{2}k} {1k} = 2\sqrt{2}$

$\sec \theta= \frac{AC} {AB} = \frac{2\sqrt{2}k}{\sqrt{7}k} = \frac{2\sqrt{2}}{\sqrt{7}}$
$\Rightarrow \frac{(cosec^{2}\theta – \sec ^{2} \theta )}{(cosec^{2}\theta + \sec ^{2} \theta )} = \begin{bmatrix} \frac{(2\sqrt{2})^{2}-(\frac{2\sqrt{2}}{\sqrt{7}})^{2}}{(2\sqrt{2})^{2} + (\frac{2\sqrt{2}}{\sqrt{7}})^{2}} \end{bmatrix}$

= $\left ( \frac{8 – \frac{8}{7}}{8 + \frac{8}{7}} \right )=\frac{\left ( \frac{48}{7} \right )}{\left ( \frac{64}{7} \right )}=\frac{48}{64} = \frac{3}{4}$

$\Rightarrow \frac{(cosec^{2}\theta – \sec ^{2} \theta )}{(cosec^{2}\theta + \sec ^{2} \theta )} = \frac{3}{4}$

Q12:

If $tan\Theta =\frac{20}{21}$, show that $\frac{\left ( 1-sin\Theta +cos\Theta \right )}{\left ( 1+sin\Theta +cos\Theta \right )}=\frac{3}{7}$

Sol:

Given:

$\tan \theta = \frac{20}{21}= \frac{20k}{21k}$

Let us draw a $\bigtriangleup ABC$ in which $\angle B = 90^{\circ} \; and \; \angle A = \theta$

By pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2} = (21k)^{2} + (20k)^{2}$
$= 441k^{2} + 400^{2}$
$= 841k^{2}$

Therefore, $AC = 29k$

$\sin \theta = \frac{BC}{AC} = \frac{20k}{29k} = \frac{20}{29}$
$\cos \theta = \frac{AB}{AC} = \frac{21k}{29k} = \frac{21}{29}$

L.H.S. = $= \frac{1- \sin \theta +\cos \theta }{1+ \sin \theta +\cos \theta } = \frac{1 – \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} = \frac{\frac{29-20+21}{29}}{\frac{29+21+21}{29}}$

= $\frac{20}{70} = \frac{3}{7}$ = R.H.S.

Q13:

$sec\Theta =\frac{5}{4},\:show\:that\:\frac{\left ( sin\Theta -2cos\Theta \right )}{tan\Theta -cot\Theta }=\frac{12}{7}$

Sol:

We have,

$sec\Theta =\frac{5}{4}$
$\Rightarrow \frac{1}{cos\Theta }=\frac{5}{4}$
$\Rightarrow cos\Theta =\frac{4}{5}$

Also,

$sin^{2}\Theta =1-cos^{2}\Theta$
$=1-\left ( \frac{4}{5} \right )^{2}$
$=1-\frac{16}{25}$
$=\frac{9}{25}$
$\Rightarrow sin\Theta =\frac{3}{5}$

Now,

$L.H.S=\frac{\left ( sin\Theta -2cos\Theta \right )}{\left ( tan\Theta -cot\Theta \right ) }$
$=\frac{\left ( sin\Theta -2cos\Theta \right )}{\left ( \frac{sin\Theta }{cos\Theta }-\frac{cos\Theta }{sin\Theta } \right )}$
$=\frac{\left ( sin\Theta -2cos\Theta \right )}{\left ( \frac{sin^{2}\Theta -cos^{2}\Theta }{sin\Theta cos\Theta } \right )}$
$=\frac{sin\Theta cos\Theta \left ( sin\Theta -2cos\Theta \right )}{sin^{2}\Theta -cos^{2}\Theta }$
$=\frac{\frac{3}{5}\times \frac{4}{5}\left ( \frac{3}{5}-2\times \frac{4}{5} \right )}{\left ( \frac{3}{5} \right )^{2}-\left ( \frac{4}{5} \right )^{2}}$
$=\frac{\frac{12}{25}\left ( \frac{3}{5}-\frac{8}{5} \right )}{\left ( \frac{9}{25}-\frac{16}{25} \right )}$
$=\frac{\frac{12}{5}\times \left ( \frac{-5}{5} \right )}{\left ( \frac{-7}{25} \right )}$
$=\frac{12}{7}$

R.H.S

Q14:

If $cot\Theta =\frac{3}{4},\:show\:that\:\sqrt{\frac{sec\Theta -cosec\Theta }{sec\Theta +cosec\Theta }}=\frac{1}{\sqrt{7}}$

Sol:

L.H.S = $\sqrt{\frac{sec\Theta -cosec\Theta }{sec\Theta +cosec\Theta }}$

$=\sqrt{\frac{\left ( \frac{1}{cos\Theta }-\frac{1}{sin\Theta } \right )}{\left ( \frac{1}{cos\Theta }+\frac{1}{sin\Theta } \right )}}$
$=\sqrt{\frac{\left ( \frac{sin\Theta -cos\Theta }{sin\Theta cos\Theta } \right )}{\left ( \frac{sin\Theta +cos\Theta }{sin\Theta cos\Theta } \right )}}$
$=\sqrt{\frac{\left ( \frac{sin\Theta -cos\Theta }{sin\Theta } \right )}{\left ( \frac{sin\Theta +cos\Theta }{sin\Theta } \right )}}$
$=\sqrt{\frac{\left ( \frac{sin\Theta }{sin\Theta }-\frac{cos\Theta }{sin\Theta } \right )}{\left ( \frac{sin\Theta }{sin\Theta }+\frac{cos\Theta }{sin\Theta } \right )}}$
$=\sqrt{\frac{1-cot\Theta }{1+cot\Theta }}$
$=\sqrt{\frac{\left ( 1-\frac{3}{4} \right )}{\left ( 1+\frac{3}{4} \right )}}$
$=\sqrt{\frac{\left ( \frac{1}{4} \right )}{\left ( \frac{7}{4} \right )}}$
$=\sqrt{\frac{1}{7}}$
$=\frac{1}{\sqrt{7}}$

= R.H.S

Q15:

If $sin\Theta =\frac{3}{4},\:show\:that\:\sqrt{\frac{cosec^{2}\Theta -cot^{2}\Theta }{sec^{2}\Theta -1}}=\frac{\sqrt{7}}{3}$

Sol:

L.H.S = $\sqrt{\frac{cosec^{2}\Theta -cot^{2}\Theta }{sec^{2}\Theta -1}}$

$=\sqrt{\frac{1}{tan^{2}\Theta }}$
$=\sqrt{cot^{2}\Theta }$
$=cot\Theta$
$=\sqrt{cosec^{2}\Theta -1 }$
$=\sqrt{\left ( \frac{1}{sin\Theta }^{2}-1 \right )}$
$=\sqrt{\left ( \frac{1}{\left ( \frac{3}{4} \right )} \right )^{2}-1}$
$=\sqrt{\left ( \frac{4}{3} \right )^{2}-1}$
$=\sqrt{\frac{16}{9}-1}$
$=\sqrt{\frac{16-9}{9}}$
$=\sqrt{\frac{7}{9}}$
$=\frac{\sqrt{7}}{3}$

= R.H.S

#### Practise This Question

Which one of the following is a hormonal disease/disorder?