# RS Aggarwal Solutions Class 10 Trigonometric Ratios

## RS Aggarwal Solutions Class 10 Chapter 5

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

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