# RS Aggarwal Class 10 Solutions Chapter 5 -Trigonometric Ratios -Ex 5A(5.1)

## RS Aggarwal Class 10 Chapter 5 -Trigonometric Ratios -Ex 5A(5.1) Solutions Free PDF

The Class 10 RS Aggarwal solutions for Chapter 5 will help students to have a deeper understanding of the concepts so that they can fetch good marks in their exams. These solutions not only make the learning process easier but also provides variety of shortcuts and tricks to solve tricky and difficult questions. Practicing RS Aggarwal Class 10 solutions will prepare them to solve any type of questions asked in the exam from any of the topics.

The solutions are prepared by our highly skilled subject experts keeping in mind the importance of it from the exam point of view. It also helps in improving the performance level of the students. Students can refer to these exercise solutions when they get stuck or have any doubts. So, if you are preparing for the Class 10 Board exam with the help of RS Aggarwal Class 10 Solutions Chapter 5 – Trigonometric Ratios then you stand a good chance of scoring better marks in the exam.

## Download PDF of RS Aggarwal Class 10 Solutions Chapter 5 – Trigonometric Ratios Ex 5A (5.1)

Q1:

If $sin\Theta =\frac{a}{b},\:show\:that\:\left ( sec\Theta +tan\Theta \right )=\sqrt{\frac{b+a}{b-a}}$

Sol:

L.H.S = $\left (sec\Theta +tan\Theta \right )$ $=\frac{1}{cos\Theta }+\frac{sin\Theta }{cos\Theta }$ $=\frac{1+sin\Theta }{cos\Theta }$ $=\frac{1+sin\Theta }{\sqrt{1-sin^{2}\Theta }}$ $=\frac{\left ( 1+\frac{a}{b} \right )}{\sqrt{1-\left ( \frac{a}{b} \right )^{2}}}$ $=\frac{\left ( \frac{1}{1}+\frac{a}{b} \right )}{\sqrt{\frac{1}{1}-\frac{a^{2}}{b^{2}}}}$ $=\frac{\left ( \frac{b+a}{b} \right )}{\sqrt{\frac{b^{2}-a^{2}}{b}}}$ $=\frac{\left ( \frac{b+a}{b} \right )}{\left ( \frac{\sqrt{b^{2}-a^{2}}}{b} \right )}$ $=\frac{\left ( b+a \right )}{\sqrt{\left ( b+a \right )\left ( b-a \right )}}$ $=\frac{\left ( b+a \right )}{\sqrt{\left ( b+a \right )}\sqrt{\left ( b-a \right )}}$ $=\frac{\sqrt{\left ( b+a \right )}}{\sqrt{\left ( b-a \right )}}$ $=\sqrt{\frac{b+a}{b-a}}$

R.H.S

Q2:

If $cos\Theta =\frac{3}{5},\:show\:that\:\frac{\left ( sin\Theta -cot\Theta \right )}{2tan\Theta }=\frac{3}{160}$

Sol:

L.H.S = $\frac{\left ( sin\Theta -cot\Theta \right )}{2tan\Theta }$ $=\frac{\left ( sin\Theta -\frac{cos\Theta }{sin\Theta } \right )}{2\left ( \frac{sin\Theta }{cos\Theta } \right )}$ $=\frac{\left ( \frac{sin^{2}\Theta -cos\Theta }{sin\Theta } \right )}{\left ( \frac{2sin\Theta }{cos\Theta } \right )}$ $=\frac{cos\Theta \left ( sin^{2}\Theta -cos\Theta \right )}{2sin^{2}\Theta }$ $=\frac{cos\Theta \left ( 1-cos^{2}\Theta -cos\Theta \right )}{2\left ( 1-cos^{2}\Theta \right )}$ $=\frac{\frac{3}{5}\left [ 1-\left ( \frac{3}{5} \right )^{2}-\frac{3}{5} \right ]}{2\left [ 1-\left ( \frac{3}{5} \right ) ^{2}\right ]}$ $=\frac{\frac{3}{5}\left ( \frac{1}{1}-\frac{9}{25}-\frac{3}{5} \right )}{2\left ( 1-\frac{9}{25} \right )}$ $=\frac{\frac{3}{5}\left ( \frac{25-9-15}{25} \right )}{2\left ( \frac{25-9}{25} \right )}$ $=\frac{\frac{3}{5}\left ( \frac{1}{25} \right )}{2\left ( \frac{16}{25} \right )}$ $=\frac{3}{5\times 2\times 16}$ $=\frac{3}{160}$

R.H.S

Q3:

If $\tan \theta = \frac{4}{3},$ show that $\sin \theta +\cos \theta =\frac{7}{5}$

Sol:

Given:

$\tan \theta =\frac{BC}{AB}= \frac{4}{3}$

Let AB = 4k and AC = 3k,

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

$AC^{2}=AB^{2}+BC^{2}$ $=(3k)^{2}+(4)k^{2}$ $=(9k^{2})+(16k^{2})$

= $=25k^{2}$

AC = 5k

Therefore, $\sin \theta= \frac{BC}{AC} = \frac{4k} {5k} = \frac{4} {5}$ $\cos \theta= \frac{AB} {AC} = \frac{3k} {5k} = \frac{3} {5}$

Therefore, $\sin \theta +\cos \theta =\left ( \frac{4}{5}+\frac{3}{5} \right )$ $Hence\; \; \sin \theta +\cos \theta =\frac{7}{5}$

Q4:

If $\tan \theta = \frac{a}{b}$, show that $\frac{a\sin \theta – b \cos \theta } {a\sin \theta + b \cos \theta} = \frac{a^{2}- b^{2}}{a^{2}+b^{2}}$

Sol:

Given:

$\tan \theta = \frac{a}{b}= \frac{ak}{bk}= \frac{BC}{AB}$

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

Pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2} = ??b^{2}k^{2} + a^{2}k^{2}$

Therefore, $AC = k \sqrt{a^{2}+b^{2}}$ $\sin \theta = \frac{BC}{AC} = \frac{ak}{k\sqrt{a^{2}+b^{2}}} = \frac{a}{\sqrt{a^{2}+b^{2}}}$ $\cos \theta = \frac{AB}{AC} = \frac{bk}{k\sqrt{a^{2}+b^{2}}} = \frac{b}{\sqrt{a^{2}+b^{2}}}$

L.H.S. = $\frac{a\sin \theta – b \cos \theta }{a\sin \theta + b \cos \theta}$ $= \frac{a \frac{a}{\sqrt{a^{2} + b^{2}}} – b. \frac{b}{\sqrt{a^{2} + b^{2}}}}{a.b} = \frac{\frac{a^{2}-b^{2}}{\sqrt{a^{2}+b^{2}}}}{\frac{a^{2} + b^{2}}{\sqrt{a^{2}+b^{2}}}} = \frac{a^{2}- b^{2}}{a^{2}+b^{2}}$

= R.H.S.

Q5:

If $3\tan \theta =4$, show that $\frac{4\cos \theta – \sin \theta }{2\cos \theta + \sin \theta } =\frac{4}{5}$.

Sol:

Given:

$3\tan \theta =4$ $\tan \theta = \frac{4}{3}$

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

By pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2}$ $= (4k)^{2} + (3k)^{2}$ $= 16k^{2} + 9k^{2} = 25k^{2}$ $AC = 5k$,

Now,

$\sin \theta= \frac{BC}{AB} = \frac{4k} {5k} = \frac{4} {5}$ $\cos \theta = \frac{AB}{AC}= \frac{3k} {5k} = \frac{3}{5}$

L.H.S. = $\frac{4\cos \theta – \sin \theta }{2\cos \theta + \sin \theta } = \frac{4\times \frac{\frac{3}{5}- \frac{4}{5}}{}}{2\times \frac{3}{4}+\frac{4}{5}}$ $\frac{12-4}{6+4}= \frac{8}{10}= \frac{4}{5}$ = R.H.S.

Q6:

If $3\cot \theta = 2$, ??show that $\frac{4\sin \theta – 3\cos \theta }{2\sin \theta +6\cos \theta }=\frac{1}{3}$

Sol:

Given:

$\cot \theta =\frac{2}{3} = \frac{2k}{3k}$

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

By pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2}$ $AC^{2} = (2k)^{2} + (3k)^{2}$ $AC^{2} = (4 k)^{2} + 9k^{2} = ??13k^{2}$ $AC = \sqrt{13k^{2}} = \sqrt{13}k$ $\sin \theta = \frac{3k}{\sqrt{13}k} = \frac{3}{\sqrt{13}}$ $\cos \theta = \frac{2k}{\sqrt{13}k} = \frac{2}{\sqrt{13}}$

L.H.S.= $= \frac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }$ $= \frac{4\times \frac{3}{\sqrt{13}} -3 \times \frac{2}{\sqrt{13}} }{2\times \frac{3}{\sqrt{13}} +6\times \frac{2}{\sqrt{13}} }$ $= \frac{\frac{12-6}{\sqrt{13}}}{\frac{6+12}{\sqrt{13}}}$ $= \frac{6}{18}=\frac{1}{3}$ = R.H.S.

Q7:

If $3cot\Theta =4,\:show\:that\:\frac{\left ( 1-tan^{2}\Theta \right )}{\left ( 1+tan^{2}\Theta \right ) }=\left ( cos^{2} \Theta -sin^{2}\Theta \right )$.

Sol:

L.H.S. = $\frac{\left ( 1-tan^{2}\Theta \right )}{\left ( 1+tan^{2}\Theta \right ) }$ $=\frac{\left ( 1-\frac{1}{cot^{2}\Theta } \right )}{\left ( 1+\frac{1}{cot^{2}\Theta } \right )}$ $=\frac{\left ( \frac{cot^{2}\Theta -1}{cot^{2}\Theta } \right )}{\left ( \frac{cot^{2}\Theta +1}{cot^{2}\Theta } \right )}$ $=\frac{cot^{2}\Theta -1}{cot^{2}\Theta +1}$ $=\frac{\left ( \frac{4}{3}\right )^{2}-1}{\left (\frac{4}{3} \right )^{2}+1}$ $\left ( As,\:3cot\Theta =4\:or\:cot\Theta =\frac{4}{3} \right )$

$=\frac{\left ( \frac{16-9}{9} \right )}{\left ( \frac{16+9}{9} \right )}$ $=\frac{\left ( \frac{7}{9} \right )}{\left ( \frac{25}{9} \right )}$ $=\frac{7}{25}$

R.H.S = $\left ( cos^{2} \Theta -sin^{2}\Theta \right )$ $=\frac{\left ( cos^{2}\Theta -sin^{2}\Theta \right )}{1}$ $=\frac{\left ( \frac{cos^{2}\Theta -sin^{2}\Theta }{sin^{2}\Theta } \right )}{\left ( \frac{1}{sin^{2}\Theta } \right )}$ $=\frac{\left ( \frac{cos^{2}\Theta }{sin^{2}\Theta }-\frac{sin^{2}\Theta }{sin^{2}\Theta } \right )}{cosec^{2}\Theta }$ $=\frac{\left ( cot^{2}-1 \right )}{\left ( cot^{2}\Theta +1 \right )}$ $=\frac{\left [ \left ( \frac{4}{3} \right )^{2}-1 \right ]}{\left [ \left ( \frac{4}{3} \right )^{2}+1 \right ]}$ $=\frac{\left ( \frac{16-9}{9} \right )}{\left ( \frac{16+9}{9} \right )}$ $=\frac{\left ( \frac{7}{9} \right )}{\left ( \frac{25}{9} \right )}$ $=\frac{7}{25}$

Since, L.H.S = R.H.S

Hence verified.

Q8:

$If\:sec\Theta =\frac{17}{8},verify\:that\:\frac{\left ( 3-4sin^{2}\Theta \right )}{4cos^{2}\Theta -3}=\frac{\left ( 3-tan^{2} \Theta \right )}{\left ( 1-3tan^{2}\Theta \right ) }$

Sol:

Given:

$\sec \theta = \frac{17}{8} = \frac{17k}{8k}$

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

By pythagoras theorem, we have

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

Therefore, $BC = \sqrt{225k^{2}} = 15k$ $\sin \theta = \frac{BC}{AC} = \frac{15k}{17k} \frac{15} {17}$ $\cos \theta = \frac{AB}{AC} = \frac{ 8k} {17k} = \frac{8}{17}$ $\tan \theta = \frac{15k} {8k} = \frac{15}{8}$

L.H.S. = $\frac{3-4\sin ^{2}\theta }{4\cos ^{2}\theta -3}= \frac{3-4\times \left ( \frac{15}{17} \right )^{2}}{4\times \left ( \frac{8}{17} \right )^{2} – 3}= \frac{3- \frac{4\times 225}{289}}{4\times \frac{64}{289}-3}$

= $\frac{3-4\sin ^{2}\theta }{4\cos ^{2}\theta -3}= \frac{3-4\times \left ( \frac{15}{17} \right )^{2}}{4\times \left ( \frac{8}{17} \right )^{2} – 3}= \frac{3- \frac{4\times 225}{289}}{4\times \frac{64}{289}-3}$

R.H.S. = $\frac{3 – \tan ^{2}\theta }{1 – 3\tan ^{2}\theta}= \frac{ 3 – \times \left ( \frac{15}{8} \right )^{2}}{1- 3\times \left ( \frac{15}{8} \right )^{2} }= \frac{3- \frac{225}{64}}{1 – 3\times \frac{225}{64}}= \frac{\frac{3\times 64- 225}{64}}{\frac{64 – 3\times 225}{64}}$ $= \frac{192-225}{64-675} = \frac{-33}{-611} = \frac{33}{611}$

Q9:

In the adjoining figure, $\angle B=90^{\circ},\angle BAC=\Theta ^{\circ}$, BC = CD = 4cm and AD = 10cm. Find $\left ( i \right )sin\Theta \:and\:\left ( ii \right )cos\Theta$

Sol:

In $\Delta ABD$,

Using Pythagoras theorem, we get

$AB=\sqrt{AD^{2}-BD^{2}}$ $=\sqrt{10^{2}-8^{2}}$ $=\sqrt{100-64}$ $=\sqrt{36}$

= 6 cm

Again,

In $\Delta ABC$,

Using Pythagoras theorem, we get

$AC=\sqrt{AB^{2}+BC^{2}}$ $=\sqrt{6^{2}+4^{2}}$ $=\sqrt{36+16}$ $=\sqrt{52}$ $=2\sqrt{13}cm$

Now,

(i) $sin\Theta =\frac{BC}{AC}$ $=\frac{4}{2\sqrt{13}}$ $=\frac{2}{\sqrt{13}}$ $=\frac{2\sqrt{13}}{13}$

(ii) $cos\Theta =\frac{AB}{AC}$ $=\frac{6}{2\sqrt{13}}$ $=\frac{3}{\sqrt{13}}$ $=\frac{3\sqrt{13}}{13}$

Q10:

In a $\Delta ABC, \angle B=90^{\circ},AB=24cm\;and\;BC=7cm.$

Find (i) sin A (ii) cos A (iii) sin C (iv) cos C

Sol:

Given:

$\bigtriangleup ABC$ in which $\angle B = 90^{\circ}$, AB= 24 cm, BC = 7 cm

By pythagoras theorem, we have

$AC^{2} = 24^{2} + 7^{2} \; cm^{2} = ( 576 + 49 ) \; cm^{2}$ $= 625 \; cm^{2}$ $= 25 \; cm^{2}$

Therefore, $AC = 25$ cm

(i) For T-ratio of $\angle A$, we have

Base = AB = 24 cm

Perpendicular = BC = 7 cm and

Hypotenuse = AC = 25 cm

$\cos A = \frac{AB}{AC}= \frac{24}{25}$ and

$\sin A= \frac{BC}{AC}= \frac{7}{25}$

(ii) For T-ratio of $\angle C$, we have

Base = BC = 7 cm

Perpendicular = AB = 24 cm and

Hypotenuse = AC = 25 cm

$\cos C = \frac{BC}{AC}= \frac{7}{25}$ and

$\sin C = \frac{AB} {AC}= \frac{24} {25}$

Q11:

In a $\Delta ABC,\angle C=90^{\circ},\angle ABC=\Theta ,BC=21$ units and AB = 29 units.

Show that $\left ( cos^{2}\Theta -sin^{2} theta\right )=\frac{41}{841}$

Sol:

Given:

$\bigtriangleup ABC$ in which $\angle B = 90^{\circ} \; \angle A = \theta$, BC = 21 units, AB = 29 units

(i) $(\cos ^{2}\theta +\sin ^{2}\theta )= \left ( \frac{21}{29} \right )^{2}+ \left ( \frac{20}{29} \right )^{2}$ $= \left ( \frac{441}{841} \right )+ \left ( \frac{400}{841} \right )$ $= \left ( \frac{841}{841} \right ) = 1$ $(\cos ^{2}\theta +\sin ^{2}\theta )= 1$

(ii) $(\cos ^{2}\theta – \sin ^{2}\theta )= \left ( \frac{21}{29} \right )^{2} – ??\left ( \frac{20}{29} \right )^{2}$ $= \left ( \frac{441}{841} \right ) – \left ( \frac{400}{841} \right )$ $= \left ( \frac{41}{841} \right ) = 1$

Hence, ??$(\cos ^{2}\theta – \sin ^{2}\theta ) = 1$

Q12:

In a $\Delta ABC,\angle B=90^{\circ},AB=12cm\:and\:BC=5cm.$

Find (i) cos A (ii) cosec A (iii) cos C (iv) cosec C

Sol:

Given:

$\bigtriangleup ABC$ in which $\angle B = 90^{\circ}$,

AB= 12 cm, BC= 5cm

By pythagoras theorem, we have

$AC^{2} = AB^{2} + BC^{2} = (12)^{2} + (5)^{2}$ $= 144k^{2} + 25^{2}$ $= 169^{2}$

Therefore, $AC = 13$

For T-ratio of $\angle A$, we have

Base = AB = 12 cm

Perpendicular = BC = 5 cm and

Hypotenuse = AC = 13 cm

Therefore, $\cos A= \frac{AB}{AC}= \frac{12}{13}$ and

Therefore, $cosec A= \frac{AC}{BC}= \frac{13}{5}$

For T-ratio of $\angle C$, we have

Base = BC = 5 cm

Perpendicular = AB = 12 cm and

Hypotenuse = AC = 13 cm

$\cos C = \frac{BC}{AC}= \frac{5}{13}$ and

$cosec C = \frac{AC} {BC}= \frac{13} {5}$

Q13:

If $sin\alpha =\frac{1}{2},prove\:that\:\left ( 3cos\alpha -4cos^{3} \alpha \right )=0$

Sol:

L.H.S = $( 3cos\alpha -4cos^{3} \alpha )$ $=cos\alpha \left ( 3-4cos^{3}\alpha \right )$ $=\sqrt{1-sin^{2}\alpha }\left [ 3-4\left ( 1-sin^{2}\alpha \right ) \right ]$ $=\sqrt{1-\left ( \frac{1}{2} \right )^{2}}\left [ 3-4\left ( 1-\left ( \frac{1}{2} \right )^{2} \right ) \right ]$ $=\sqrt{\frac{1}{1}-\frac{1}{4}}\left [ 3-4\left ( \frac{1}{1}-\frac{1}{4} \right ) \right ]$ $=\sqrt{\frac{3}{4}}\left [ 3-4\left ( \frac{3}{4} \right ) \right ]$ $=\sqrt{\frac{3}{4}}\left [ 3-3 \right ]$ $=\sqrt{\frac{3}{4}}\left [ 0 \right ]$

= 0

= R. H. S

Q14:

In a $\Delta ABC,\angle B=90^{\circ}\:and\:tanA=\frac{1}{\sqrt{3}}.\:\:Prove\:that\:\left ( i \right )sinA.cosC+cosA.sinC=1\:\:\left ( ii \right )cosA.cosC-sinA.sinC=0$

Sol:

In $\Delta ABC,\angle B=90^{\circ}$,

As, tanA = $\frac{1}{\sqrt{3}}$ $=\frac{BC}{AB}=\frac{1}{\sqrt{3}}$

Let BC = x and AB = $x\sqrt{3}$

Using Pythagoras theorem, we get

$AC=\sqrt{AB^{2}+BC^{2}}$ $=\sqrt{\left ( X\sqrt{3} \right )^{2}+x^{2}}$ $=\sqrt{3x^{2}+x^{2}}$ $=\sqrt{4x^{2}}$

= 2x

Now,

(i) L.H.S = sinA.cosC + cosA.sinC

$=\frac{BC}{AC}.\frac{BC}{AC}+\frac{AB}{AC}.\frac{AB}{AC}$ $=\left ( \frac{BC}{AC} \right )^{2}+\left ( \frac{AB}{AC} \right )^{2}$ $=\left ( \frac{x}{2x} \right )^{2}+\left ( \frac{x\sqrt{3}}{2x} \right )^{2}$ $=\frac{1}{4}+\frac{3}{4}$ $=\frac{4}{4}$

= 1

= R.H.S

(ii) L.H.S = cosA.cosC ??? sinA.sinC

$=\frac{AB}{AC}.\frac{BC}{AC}-\frac{BC}{AC}.\frac{AB}{AC}$ $=\frac{x\sqrt{3}}{2x}.\frac{x}{2x}-\frac{x}{2x}.\frac{x\sqrt{3}}{2x}$ $=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

= 0

= R.H.S

Q15:

If $\angle A\:and\:\angle B$ are acute angles such that sinA = sinB then prove that $\angle A=\angle B$.

Sol:

In $\Delta ABC,\angle B=90^{\circ}$,

sinA = $\frac{BC}{AB}$ and

sinB = $\frac{AC}{AB}$

As, sinA = sinB

$\Rightarrow \frac{BC}{AB}=\frac{AC}{AB}$

So, $\angle A=\angle B$ (Angles opposite to each sides are equal)

Q16:

If $\angle A\:and\:\angle B$ are acute angles such that tanA = tanB then prove that $\angle A=\angle B$.

Sol:

In $\Delta ABC,\angle C=90^{\circ}$,

tanA = $\frac{BC}{AC}$ and

tanB = $\frac{AC}{BC}$

As, tanA = tanB

$\Rightarrow \frac{BC}{AC}=\frac{AC}{BC}$ $\Rightarrow BC^{2}=AC^{2}$

=> BC = AC

So, $\angle A=\angle B$ (Angles opposite to each sides are equal)

Q17:

In a right $\Delta ABC$, right-angled at B, if tanA = 1 then verify that 2sinA.cosA = 1.

Sol:

We have,

tanA = 1

=> $\Delta ABC$ = 1

=>sinA = cosA

=>sinA – cosA = 0

Squaring both sides, we get

$\left ( sinA-cosA \right )^{2}=0$ $\Rightarrow sin^{2}A+cosA^{2}-2sinA.cosA=0$

=> 1- 2sinA.cosA =0

Therefore, $2sinA.cosA=1$

Q18:

In the figure of $\Delta PQR,\angle P=\Theta ^{\circ}\:and\:\angle R=\phi ^{\circ}.$

Find:

(i) $\left ( \sqrt{x+1} \right )cot\phi$

(ii) $\left ( \sqrt{x^{3}+x^{2}} \right )tan\Theta$

(iii) $cos\Theta$

Sol:

In $\Delta PQR,\angle Q=90^{\circ}$,

Using Pythagoras theorem, we get

$PQ=\sqrt{PR^{2}-QR^{2}}$ $=\sqrt{\left ( x+2 \right )^{2}-x^{2}}$ $=\sqrt{x^{2}+4x+4-x^{2}}$ $=\sqrt{4\left ( x+1 \right )}$ $=2\sqrt{x+1}$

Now,

(i)$\left ( \sqrt{x+1} \right )cot\phi$ $=\left ( \sqrt{x+1} \right )\times \frac{QR}{PQ}$ $=\left ( \sqrt{x+1} \right )\times \frac{x}{2\sqrt{x+1}}$ $=\frac{x}{2}$

(ii)$\left ( \sqrt{x^{3}+x^{2}} \right )tan\Theta$ $=\left ( \sqrt{x^{2}\left ( x+1 \right )} \right )\times \frac{QR}{PQ}$ $=x\sqrt{\left ( x+1 \right )}\times \frac{x}{2\sqrt{x+1}}$ $=\frac{x^{2}}{2}$

(iii)$cos\Theta$ $=\frac{PQ}{PR}$ $=\frac{2\sqrt{x+1}}{\left ( x+2 \right )}$

Q19:

If x = cosecA + cosA and y = cosecA ??? cosA then prove that

$\left ( \frac{2}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}-1=0$

Sol:

L.H.S = $\left ( \frac{2}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}-1$ $=\left [ \frac{2}{\left ( cosecA+cosA \right )+\left ( cosecA-cosA \right )} \right ]^{2}+\left [ \frac{\left ( cosecA+cosA \right )-\left ( cosecA-cosA \right )}{2} \right ]^{2}-1$ $=\left [ \frac{2}{cosecA+cosA+cosecA-cosA} \right ]^{2}+\left [ \frac{cosecA+cosA-cosecA+cosA}{2} \right ]^{2}-1$ $=\left [ \frac{2}{2cosecA} \right ]^{2}+\left [ \frac{2cosA}{2} \right ]^{2}-1$ $=\left [ \frac{1}{cosecA} \right ]^{2}+\left [ cosA \right ]^{2}-1$ $=\left [ sinA \right ]^{2}+\left [ cosA \right ]^{2}-1$ $=sin^{2}A+cos^{2}A-1$

= 1-1 = 0

= 0

= R.H.S

Q20:

If x = cot A + cos A and y = cot A ??? cos A, prove that

$\left ( \frac{x-y}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}=1$

Sol:

L.H.S = $\left ( \frac{x-y}{x+y} \right )^{2}+\left ( \frac{x-y}{2} \right )^{2}$ $=\left [ \frac{\left ( cotA+cosA \right )-\left ( cotA-cosA \right )}{\left ( cotA+cosA \right )+\left ( cotA-cosA \right )} \right ]^{2}+\left [ \frac{\left ( cotA+cosA \right )-\left ( cotA-cosA \right )}{2} \right ]^{2}$ $=\left [ \frac{cotA+cosA-cotA+cosA}{cotA+cosA+cotA-cosA} \right ]+\left [ \frac{cotA+cosA-cotA+cosA}{2} \right ]^{2}$ $=\left [ \frac{2cosA}{2cotA} \right ]^{2}+\left [ \frac{2cosA}{2} \right ]^{2}$ $=\left [ \frac{cosA}{\left ( \frac{cosA}{sinA} \right )} \right ]^{2}+\left [ cosA \right ]^{2}$ $=\left [ \frac{sinA.cosA}{cosA} \right ]^{2}+\left [ cosA \right ]^{2}$ $=\left [ sinA \right ]^{2}+\left [ cosA \right ]^{2}$ $=sin^{2}A+cos^{2}A$

= 1

= R.H.S

### Key Features of RS Aggarwal Class 10 Solutions Chapter 5 – Trigonometric Ratios Ex 5A (5.1)

• It boosts up the students confidence level while attempting the final question paper.
• It is considered as one of the perfect study material from the exam point of view.
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#### Practise This Question

D and E are points on the sides AB and AC respectively of a ABC  such that DE || BC. Which of the following statement is true?

(i)   ADE ~  ABC