# RS Aggarwal Solutions Class 10 Chapter 6 T Ratios Of Some Particular Angles

## RS Aggarwal Class 10 Chapter 6 T Ratios Of Some Particular Angles Solutions Free PDF

The T-ratios of some particular angles stands for the trigonometric ratios of a few select angles. Some of the particular angles are 0, 30, 45, 60 and 90-degree angles. Understanding the trigonometric ratios of these angles is incredibly important, as it helps the students to solve trigonometric problems quite easily. It is possible for the students to construct an entire tabular column, of all the trigonometric ratios of different angles by simply being able to calculate all the angles of a particular ratio.

For example, if they can deduce Sin 0Â°, Sin 30Â°, Sin 45Â°, Sin 60Â°, and Sin 90Â° then they can easily find out all the other trigonometric ratios as well.

## Download PDF of RS Aggarwal Class 10 Chapter 6-Â T-Ratios of Some Particular Angles

Class 10 is considered the turning point in their academic career. To score good marks in a subject like maths students needs lots of practice. So, in a way to help them, we at BYJU’S provide RS Aggarwal Class 10 solutions to practice, which covers the entire maths syllabus. It is one of the best reference material for preparation purpose.

Evaluate each of the following:

Question.1: sin 60o cos 30o + cos 60o sin 30o

Solution:

On substituting the values of various T-ratios, we get:

sin 60o cos 30o + cos 60o sin 30o

= $\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$

Question.2: cos 60o cos 30o â€“ sin 60o sin 30o

Solution:

On substituting the value of various T-ratios, we get:

cos 60o cos 30o â€“ sin 60o sin 30o

= $\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$

Question.3: cos 45o cos 30o + sin 45o sin 30o

Solution:

On substituting the value of various T-ratios, we get:

cos 45o cos 30o + sin 45o sin 30o

= $\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Question.4: $\frac{sin30^{\circ}}{cos45^{\circ}}+\frac{cot45^{\circ}}{sec60^{\circ}}-\frac{sin60^{\circ}}{tan45^{\circ}}+\frac{cos30^{\circ}}{sin90^{\circ}}$

Solution:

On substituting the values of various T-ratios, we get:

$\frac{sin30^{\circ}}{cos45^{\circ}}+\frac{cot45^{\circ}}{sec60^{\circ}}-\frac{sin60^{\circ}}{tan45^{\circ}}+\frac{cos30^{\circ}}{sin90^{\circ}}$

= $\frac{\left ( \frac{1}{2} \right )}{\left ( \frac{1}{\sqrt{2}} \right )}+\frac{1}{\left ( \frac{2}{1} \right )}-\frac{\left ( \frac{\sqrt{3}}{2} \right )}{1}-\frac{\left ( \frac{\sqrt{3}}{2} \right )}{1}$

= $\frac{\sqrt{2}}{2}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=\frac{\sqrt{2}+1-\sqrt{3}-\sqrt{3}}{2}$

= $\left ( \frac{\sqrt{2}+1-2\sqrt{3}}{2} \right )$

Question.5: $\frac{5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2}45^{\circ}}{sin^{2}30^{\circ}+cos^{2}30^{\circ}}$

Solution:

On substituting the value of various T-ratios, we get:

$\frac{5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2}45^{\circ}}{sin^{2}30^{\circ}+cos^{2}30^{\circ}}$

= $\frac{5\times \left ( \frac{1}{2} \right )^{2}+4\times \left ( \frac{2}{\sqrt{3}} \right )^{2}-(1)^{2}}{\left ( \frac{1}{2} \right )^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}}$

= $\frac{5\times \frac{1}{4}+4\times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}$

= $\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}$

= $\frac{\frac{15+64-12}{12}}{\frac{4}{4}}$ = $\frac{67}{12}$

Question.6: 2cos2 60o + 3sin2 45o â€“ 3sin2 30o + 2cos2 90o

Solution:

On substituting the value of various T-ratios, we get:

2cos2 60o + 3sin2 45o â€“ 3sin2 30o + 2cos2 90o

= $2\times \left ( \frac{1}{2} \right )^{2}+3\times \left ( \frac{1}{\sqrt{2}} \right )^{2}-3\times \left ( \frac{1}{2} \right )^{2}+2(0)^{2}$

= $\frac{1}{2}+\frac{3}{2}-\frac{3}{4}=\frac{2+6-3}{4}=\frac{5}{4}$

Question.7: cot2 30o â€“ 2cos2 30o â€“ $\frac{3}{4}$sec2 45o + $\frac{1}{4}$cosec2 30o

Solution:

On substituting the value of various T-ratios, we get:

cot2 30o â€“ 2cos2 30o â€“ $\frac{3}{4}$sec2 45o + $\frac{1}{4}$cosec2 30o

= $(\sqrt{3})^{2}-2\times \left ( \frac{\sqrt{3}}{2} \right )^{2}-\frac{3}{4}\times \left ( \frac{\sqrt{2}}{1} \right )^{2}+\frac{1}{4}\times (2)^{2}$

= $3-2\times \frac{3}{4}-\frac{3}{4}\times 2+\frac{1}{4}\times 4$

= $3-\frac{3}{2}-\frac{3}{2}+1$ = $\frac{6-3-3+2}{2}=\frac{2}{2}=1$

Question.8: (sin2 30o + 4cot2 45o â€“ sec2 60o) (cosec2 45o sec2 30o)

Solution:

On substituting the value of various T-ratios, we get:

(sin2 30o + 4cot2 45o â€“ sec2 60o)(cosec2 45o sec2 30o)

= $\left [ \left ( \frac{1}{2} \right )^{2}+4\times (1)^{2}-(2)^{2} \right ]\left [ (\sqrt{2})^{2}\times \left ( \frac{2}{\sqrt{3}} \right )^{2} \right ]$

= $\left ( \frac{1}{4}+4-4 \right )\left ( 2\times \frac{4}{3} \right )$

= $\frac{1}{4}\times \frac{8}{3}=\frac{2}{3}$

Question.9: $\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}30^{\circ}}-2cos^{2}45^{\circ}-sin^{2}0^{\circ}$

Solution:

On substituting the value of various T-ratios, we get:

$\frac{4}{cot^{2}30^{\circ}}+\frac{1}{sin^{2}30^{\circ}}-2cos^{2}45^{\circ}-sin^{2}0^{\circ}$

= $\frac{4}{(\sqrt{3})^{2}}+\frac{1}{\left ( \frac{1}{2} \right )^{2}}-2\times \left ( \frac{1}{\sqrt{2}} \right )^{2}-0$

= $\frac{4}{3}+\frac{4}{1}-\frac{2}{2}-0$

= $\frac{8+24-6-0}{6}$

= $\frac{26}{6}=\frac{13}{3}$

Question.10: Show that:

(i) $\frac{1-sin60^{\circ}}{cos60^{\circ}}=\frac{tan60^{\circ}-1}{tan60^{\circ}+1}$

(ii) $\frac{cos30^{\circ}+sin60^{\circ}}{1+sin30^{\circ}+cos60^{\circ}}=cos30^{\circ}$

Solution:

(i) L.H.S. = $\frac{1-sin60^{\circ}}{cos60^{\circ}}$ = $\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{2-\sqrt{3}}{1}$

R.H.S. = $\frac{tan60^{\circ}-1}{tan60^{\circ}+1}$

= $\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

= $\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-(1)^{2}}$

= $\frac{3+1-2\sqrt{3}}{3-1}$

= $\frac{4-2\sqrt{3}}{2}$ = $\frac{2(2-\sqrt{3})}{2}$ = $(2-\sqrt{3})$

L.H.S. = R.H.S.

Hence, $\frac{1-sin60^{\circ}}{cos60^{\circ}}=\frac{tan60^{\circ}-1}{tan60^{\circ}+1}$

(ii) L.H.S. = $\frac{cos30^{\circ}+sin60^{\circ}}{1+sin30^{\circ}+cos60^{\circ}}=\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}$

R.H.S. = cos 30o = $\frac{\sqrt{3}}{2}$

L.H.S. = R.H.S.

Hence, $\frac{cos30^{\circ}+sin60^{\circ}}{1+sin30^{\circ}+cos60^{\circ}}=cos30^{\circ}$

Question.11: Verify each of the following:

(i) sin 60o cos 30o â€“ cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2sin 30o cos 30o = sin 60o

(iv) 2sin 45o cos 45o = sin 90o

Solution:

(i) L.H.S. = sin 60o cos 30o â€“ cos 60o sin 30o

= $\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$

R.H.S. = sin 30o = $\frac{1}{2}$

R.H.S. = L.H.S.

Hence, sin 60o cos 30o â€“ cos 60o sin 30o = sin 30o

(ii) L.H.S. = cos 60o cos 30o + sin 60o sin 30o

= $\frac{1}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$

R.H.S. = cos 30o = $\frac{\sqrt{3}}{2}$

R.H.S. = L.H.S.

Hence, cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) L.H.S. = 2sin 30o cos 30o

= $2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$

R.H.S. = sin 60o = $\frac{\sqrt{3}}{2}$

R.H.S. = L.H.S.

Hence, 2sin 30o cos 30o = sin 60o

(iv) L.H.S. = 2sin 45o cos 45o

= $2\times \frac{1}{\sqrt{2}}\times \frac{\sqrt{1}}{\sqrt{2}}=1$

R.H.S. = sin 90o = 1

R.H.S. = L.H.S.

Hence, 2sin 45o cos 45o = sin 90o

Question.12: If A = 45o, verify that:

(i) sin 2A = 2sin A cos A

(ii) cos 2A = 2cos2 A â€“ 1 = 1 â€“ 2sin2 A

Solution:

A = 45o $\Rightarrow$ 2A = 90o

(i) sin 2A = sin 90o = 1

$Therefore,$ 2sin A cos A = 2sin 45o cos 45o = $2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=1$

Therefore, sin 2A = 2sin A cos A

(ii) cos 2A = cos 90o = 0

2cos2 A â€“ 1 = 2cos2 45o â€“ 1 = $2\left ( \frac{1}{\sqrt{2}} \right )^{2}-1=1-1=0$

1 â€“ 2sin2 A = 1 â€“ 2sin2 45o = $1-2\times \left ( \frac{1}{\sqrt{2}} \right )^{2}=1-1=0$

$Therefore,$ cos 2A = 2cos2 A â€“ 1 = 1 â€“ 2sin2 A

Question.13: If A = 30o, verify that:

(i) $sin2A=\frac{2tanA}{1+tan^{2}A}$

(ii) $cos2A=\frac{1-tan^{2}A}{1+tan^{2}A}$

(iii) $tan2A=\frac{2tanA}{1-tan^{2}A}$

Solution:

A = 30o $\Rightarrow$ 2A = 60o

(i) sin 2A = sin 60o = $\frac{\sqrt{3}}{2}$

Also $\frac{2tanA}{1+tan^{2}A}=\frac{2tan30^{\circ}}{1+tan^{2}30^{\circ}}=\frac{2\times \frac{1}{\sqrt{3}}}{1+\left ( \frac{1}{\sqrt{3}} \right )^{2}}=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}$

= $\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2}$

Hence, $sin2A=\frac{2tanA}{1+tan^{2}A}$

(ii) cos 2A = cos 60o = $\frac{1}{2}$

Also, $\frac{1-tan^{2}A}{1+tan^{2}A}=\frac{1-tan^{2}30^{\circ}}{1+tan^{2}30^{\circ}}=\frac{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}}{1+\left ( \frac{1}{\sqrt{3}} \right )^{2}}$

= $\frac{\left ( 1-\frac{1}{3} \right )}{\left ( 1+\frac{1}{3} \right )}=\frac{\left ( \frac{2}{3} \right )}{\left ( \frac{4}{3} \right )}=\left ( \frac{2}{3}\times \frac{3}{4} \right )=\frac{1}{2}$

Hence, $cos2A=\frac{1-tan^{2}A}{1+tan^{2}A}$

(iii) tan 2A = tan 60o = $\sqrt{3}$

Also, $\frac{2tanA}{1-tan^{2}A}=\frac{2tan30^{\circ}}{1-tan^{2}30^{\circ}}=\frac{2\times \frac{1}{\sqrt{3}}}{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}}=\frac{2\times \frac{1}{\sqrt{3}}}{1-\frac{1}{3}}$

= $\frac{\left ( \frac{2}{\sqrt{3}} \right )}{\left ( \frac{2}{3} \right )}=\left ( \frac{2}{\sqrt{3}}\times \frac{3}{2} \right )=\sqrt{3}$

Hence, $tan2A=\frac{2tanA}{1-tan^{2}A}$

Question.14: If A = 60o and B = 30o, verify that:

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B â€“ sin A sin B

Solution:

A = 60o and B = 30o, (A + B) = (60o + 30o) = 90o

(i) L.H.S. = sin (A + B) = sin 90o = 1

R.H.S. = sin A cos B + cos A sin B

= sin 60o cos 30o + cos 60o sin 30o

= $\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$

Hence, sin (A + B) = sin A cos B + cos A sin B

(ii) L.H.S. = cos (A + B) = cos 90o = 0

R.H.S. = cos A cos B â€“ sin A sin B

= cos 60o cos 30o â€“ sin 60o sin 30o

= $\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$

Hence, cos (A + B) = cos A cos B â€“ sin A sin B

Question.15: If A = 60o and B = 30o, verify that:

(i) sin (A â€“ B) = sin A cos B â€“ cos A sin B

(ii) cos (A â€“ B) = cos A cos B + sin A sin B

(iii) tan (A â€“ B) = $\frac{tanA-tanB}{1+tanA.tanB}$

Solution:

A = 60o and B = 30o, (A â€“ B) = (60o â€“ 30o) = 30o

(i) L.H.S. = sin (A â€“ B) = sin 30o = $\frac{1}{2}$

R.H.S. = sin A cos B â€“ cos A sin B

= sin 60o cos 30o â€“ cos 60o sin 30o

= $\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}$

= $\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$

Hence, sin (A â€“ B) = sin A cos B â€“ cos A sin B

(ii) L.H.S. = cos (A â€“ B) = cos 30o = $\frac{\sqrt{3}}{2}$

R.H.S. = cos A cos B + sin A sin B

= cos 60o cos 30o + sin 60o sin 30o

= $\frac{1}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2}=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$

Hence, cos (A â€“ B) = cos A cos B + sin A sin B

(iii) L.H.S. = tan (A â€“ B) = tan 30o = $\frac{1}{\sqrt{3}}$

R.H.S. = $\frac{tanA-tanB}{1+tanA.tanB}$

= $\frac{tan60^{\circ}-tan30^{\circ}}{1+tan60^{\circ}.tan30^{\circ}}$

= $\frac{\sqrt{3-\frac{1}{\sqrt{3}}}}{1+\sqrt{3}\times \frac{1}{\sqrt{3}}}=\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}$

Hence, tan (A â€“ B) = $\frac{tanA-tanB}{1+tanA.tanB}$

Question.16: If A and B are acute angles such that tan A = $\frac{1}{3}$, tan B = $\frac{1}{2}$ and tan (A + B) = $\frac{tanA-tanB}{1-tanA.tanB}$, show that A + B = 45o.

Solution:

tan (A + B) = $\frac{tanA+tanB}{1-tanA.tanB}$

tan (A + B) = $\frac{\left ( \frac{1}{3}+\frac{1}{2} \right )}{1-\frac{1}{3}\times \frac{1}{2}}$ [$Since, tanA=\frac{1}{3},tanB=\frac{1}{2}$]

= $\frac{\frac{5}{6}}{\frac{5}{6}}=\frac{5}{6}\times \frac{6}{5}=1$

tan (A + B) = 1 $\Rightarrow$ tan (A + B) = tan 45o

Hence, (A + B) = 45o

Question.17: Using the formula, tan 2A = $\frac{2tanA}{1-tan^{2}A}$, find the value of tan 60o, it being given that tan 30o = $\frac{1}{\sqrt{3}}$.

Solution:

Given: tan 2A = $\frac{2tanA}{1-tan^{2}A}$ and tan 30o = $\frac{1}{\sqrt{3}}$

Let A = 30o $\Rightarrow$ 2A = 60o

$\Rightarrow$ tan 60o = tan 2A = $\frac{2tanA}{1-tan^{2}A}$

= $\frac{2\times \frac{1}{\sqrt{3}}}{1-\left ( \frac{1}{\sqrt{3}} \right )^{2}}$

= $\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$

= $\frac{2}{\sqrt{3}}\times \frac{3}{2}=\sqrt{3}$

Hence, tan 60o = $\sqrt{3}$

Question.18: Using the formula, cos A = $\sqrt{\frac{1+cos2A}{2}}$, find the value of cos 30o, it being given that cos 60o = $\frac{1}{2}$.

Solution:

Given: cos A = $\sqrt{\frac{1+cos2A}{2}}$ and cos 60o = $\frac{1}{2}$

Let, A = 30o $\Rightarrow$ 2A = 60o

$\Rightarrow$ cos 30o = cos A = $\sqrt{\frac{1+cos2A}{2}}$

= $\sqrt{\frac{1+\frac{1}{2}}{2}}=\sqrt{\frac{\frac{3}{2}}{2}}=\sqrt{\frac{3}{2}\times \frac{1}{2}}$

= $\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

Hence, cos 30o = $\frac{\sqrt{3}}{2}$

Question.19: Using the formula, sin A = $\sqrt{\frac{1-cos2A}{2}}$, find the value of sin 30o, it being given that cos 60o = $\frac{1}{2}$.

Solution:

Given: sin A = $\sqrt{\frac{1-cos2A}{2}}$ and cos 60o = $\frac{1}{2}$

Let, A = 30o $\Rightarrow$ 2A = 60o

$\Rightarrow$ sin 30o = sin A = $\sqrt{\frac{1-cos2A}{2}}$

= $\sqrt{\frac{1-\frac{1}{2}}{2}}=\sqrt{\frac{\frac{1}{2}}{2}}$

= $\sqrt{\frac{1}{4}}=\frac{1}{2}$

Hence, sin 30o = $\frac{1}{2}$

Question.20: In the adjoining figure, $\Delta$ABC is a right-angled triangle in which $\angle$A = 30o and AC = 20 cm. Find

(i) BC,

(ii) AB

Solution:

From right angled $\Delta$ABC,

We have $\frac{BC}{AC}$ = sin 30o

$\Rightarrow$ $\frac{BC}{20}=\frac{1}{2}\Rightarrow BC=10cm$

By Pythagoras theorem,

(AB)2 = (AC)2 â€“ (BC)2

$\Rightarrow$ AB = $\sqrt{(AC)^{2}-(BC)^{2}}$

$\Rightarrow$ AB = $\sqrt{(20)^{2}-(10)^{2}}$

$\Rightarrow$ AB = $\sqrt{300}=10\sqrt{3}cm$

Hence, BC = 10 cm and AB = $10\sqrt{3}cm$

Question.21: In the adjoining figure, $\Delta$ABC is a right-angled triangle at B and $\angle$A = 30o. If BC = 6 cm, find

(i) AB,

(ii) AC

Solution:

We have $\frac{BC}{AC}$ = sin 30o

$\Rightarrow$ $\frac{6}{AC}=\frac{1}{2}$

$\Rightarrow$ AC = 12 cm

By Pythagoras theorem,

(AB)2 = (AC)2 â€“ (BC)2

$\Rightarrow$ AB = $\sqrt{(AC)^{2}-(BC)^{2}}$

$\Rightarrow$ AB = $\sqrt{(12)^{2}-(6)^{2}}$

$\Rightarrow$ AB = $\sqrt{144-36}$

$\Rightarrow$ AB = $\sqrt{108}=6\sqrt{3}cm$

Hence, AB = $6\sqrt{3}cm$ and AC = 12 cm

Question.22: In the adjoining figure, $\Delta$ABC is a right-angled triangle at B and $\angle$A = 45o. If AC = $3\sqrt{2}$ cm, find

(i) BC,

(ii) AB.

Solution:

(i) $\frac{BC}{AC}$ = sin 45o

$\Rightarrow$ $\frac{BC}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$

$\Rightarrow$ BC = 3

(ii) By Pythagoras theorem:

(AB)2 = (AC)2 â€“ (BC)2 = $\sqrt{(3\sqrt{2})^{2}-(3)^{2}}$

$\Rightarrow$ AB = $\sqrt{18-9}=\sqrt{9}=3cm$

Hence, (i) BC = 3 cm and (ii) AB = 3 cm

Question.23: If sin (A + B) = 1 and cos (A â€“ B) = 1, 0o $\leq$ (A + B) $\leq$ 90o and A > B then find A and B.

Solution:

sin (A + B) = 1 $\Rightarrow$ sin (A + B) = sin 90o

$\Rightarrow$ A + B = 90o – – – – – – – (1)

cos (A â€“ B) = 1 $\Rightarrow$ cos (A â€“ B) = cos 0o

$\Rightarrow$ A â€“ B = 0o – – – – – – – – – (2)

Adding (1) and (2), we get 2A = 90o $\Rightarrow$ A = 45o

Putting A = 45o in (1) we get

45o + B = 90o $\Rightarrow$ B = 45o

Hence, A = 45o and B = 45o

Question.24: If sin (A â€“ B) = $\frac{1}{2}$ and cos (A + B) = $\frac{1}{2}$, 0o < (A + B) < 90o and A > B then find A and B.

Solution:

sin (A â€“ B) = $\frac{1}{2}$ $\Rightarrow$ sin (A â€“ B) = sin 30o

$\Rightarrow$ A â€“ B = 30o – – – – – – – (1)

cos (A + B) = $\frac{1}{2}$ $\Rightarrow$ cos (A + B) = cos 60o

$\Rightarrow$ A + B = 60o – – – – – – – – – (2)

Solving (1) and (2), we get 2A = 90o $\Rightarrow$ A = 45o

Putting A = 45o in (1) we get

45o â€“ B = 30o $\Rightarrow$ B = 15o

Hence, A = 45o and B = 15o

Question.25: If tan (A â€“ B) = $\frac{1}{\sqrt{3}}$ and tan (A + B) = $\sqrt{3}$, 0o < (A + B) < 90o and A > B then find A and B.

Solution:

tan (A â€“ B) = $\frac{1}{\sqrt{3}}$ $\Rightarrow$ tan (A â€“ B) = tan 30o

$\Rightarrow$ A â€“ B = 30o – – – – – – – (1)

tan (A + B) = $\sqrt{3}$ $\Rightarrow$ tan (A + B) = tan 60o

$\Rightarrow$ A + B = 60o – – – – – – – – (2) [tan 60o = $\sqrt{3}$]

Solving (1) and (2), we get 2A = 90o $\Rightarrow$ A = 45o

Putting A = 45o in (1) we get

45o â€“ B = 30o $\Rightarrow$ B = 15o

Hence, A = 45o and B = 15o

Question.26: If 3x = cosec $\theta$ and $\frac{3}{x}$ = cot $\theta$, find the value of $3\left ( x^{2}-\frac{1}{x^{2}} \right )$.

Solution:

$3(x^{2} – \frac{1}{x^{2}})$

$= \frac{9}{3}(x^{2} – \frac{1}{x^{2}})$

$= \frac{1}{3}(9x^{2} – \frac{9}{x^{2}})$

$= \frac{1}{3}\left [ (3x)^{2} – \left ( \frac{3}{x} \right )^{2} \right ]$

$= \frac{1}{3}\left [ (cosec \; \theta)^{2} – \left ( \cot \theta \right )^{2} \right ]$

$= \frac{1}{3}\left [ (cosec ^{2}\; \theta – \cot ^{2} \theta ) \right ]$

$= \frac{1}{3} (1)$ = $\frac{1}{3}$

Question.27: If sin (A + B) = sin A cos B + cos A sin B and cos (A â€“ B) = cos A cos B + sin A sin B, find the values of

(i) sin 75o and

(ii) cos 15o

Solution:

Let A = $45^{\circ}$ and B = $30^{\circ}$

(i) As, $\sin (A + B) = \sin A \cos B + \cos A \sin B$

$\Rightarrow \sin (45^{\circ} + 30 ^{\circ}) = \sin 45^{\circ} \cos 30 ^{\circ} + \cos 45^{\circ} \sin 30 ^{\circ}$

$\Rightarrow \sin (75 ^{\circ}) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}$

$\Rightarrow \sin (75 ^{\circ}) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2 \sqrt{2}}$

$Therefore, \sin (75 ^{\circ}) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

(ii) As, $\cos (A – B) = \cos A \cos B + \sin A \sin B$

$\Rightarrow \cos (45^{\circ} – 30 ^{\circ}) = \cos 45^{\circ} \cos 30 ^{\circ} + \sin 45^{\circ} \sin 30 ^{\circ}$

$\Rightarrow \cos (15 ^{\circ}) = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2}$

$\Rightarrow \cos (15 ^{\circ}) = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2 \sqrt{2}}$

$Therefore, \cos (75 ^{\circ}) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

### RS Aggarwal Class 10 Solutions Chapter 6 â€“ T-Ratios of Some Particular Angles

The solutions of RS Aggarwal Class 10 Maths are considered as extremely helpful resource for exam preparation. It provides ample number of ways to solve a particular question and explains difficult and tricky questions in a simple way. The students of Class 10 should have a good grasp of theÂ RS Aggarwal Maths Solutions for Class 10 so that they can attempt the question paper confidently. It also improves their performance, weaknesses and also helps to strengthen their knowledge.

#### Practise This Question

State true or false.
In  ABC, sin(A+B2)=cos(C2)