# RS Aggarwal Class 10 Solutions Chapter 5 - Trigonometric Ratios

## RS Aggarwal Class 10 Chapter 5 - Trigonometric Ratios Solutions Free PDF

The trigonometric ratios are the basic ratios that are found in a right-angled triangle, by using the hypotenuse and the two shorter sides of a triangle. These ratios are essential for future calculations, so understanding them is vital for all students. Using the Pythagorean theorem of a right-angled triangle, we can obtain a relationship between the sides and the angles of a triangle. The six major trigonometric ratios are:

1. Sine
2. Cosine
3. Tangent
4. Cosecant
5. Secant
6. Cotangent

## Download PDF of RS Aggarwal Class 10 Chapter 5- Trigonometric Ratios

The RS Aggarwal maths solutions are the most beneficial resources to cover the entire Class 10 maths syllabus in an efficient way. Class 10 Maths RS Aggarwal Solutions for Trigonometric Ratios are created by subject experts in a simple language for better understanding of the students.

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

### RS Aggarwal Class 10 Solutions Chapter 5 – Trigonometric Ratios

The Chapter 5 solutions of RS Aggarwal Class 10 will be very helpful for you in case of clearing your doubts quickly and learning the most easy and convinient way to write paper during your examinations. RS Aggarwal Solutions for Trigonometric Ratios provided here are solved in stepwise way with appropriate description. It contains all the necessary information and guidelines that a student requires to score well in their examination. The RS Aggarwal Class 10 solutions is the best resource for students to practice variety of questions from each and every topic.

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In the given figure, ABC PQR. Then, Area of ABCArea of PQR= _____