RS Aggarwal 2020 Solutions for Class 10 Math Chapter 11 - T Ratios Of Some Particular Angles

Question 27:

Question 28:

$$\begin{gathered} 3\left(x^2-\frac{1}{x^2}\right) \\ =\frac{9}{3}\left(x^2-\frac{1}{x^2}\right) \\ =\frac{1}{3}\left(9x^2-\frac{9}{x^2}\right) \\ =\frac{1}{3}\left[{\left(3x\right)}^2-{\left(\frac{3}{x}\right)}^2\right] \end{gathered}$$
$$\begin{gathered} =\frac{1}{3}\left[{\left(\mathrm{cosec}\theta \right)}^2-{\left(\cot \theta \right)}^2\right] \\ =\frac{1}{3}\left({\mathrm{cosec}}^2\theta -{\cot }^2\theta \right) \\ =\frac{1}{3}\left(1\right) \\ =\frac{1}{3} \end{gathered}$$

Question 29:

From the given right-angled triangle, we have:
$$\begin{gathered} \frac{BC}{AC} = \sin {30}^{\mathrm{o}} \\ \Rightarrow \frac{BC}{20} = \frac{1}{2} \\ \Rightarrow BC = \frac{20}{2} = 10 \mathrm{cm} \\ \mathrm{Also}, \frac{AB}{AC} = \cos {30}^{\mathrm{o}} \\ \Rightarrow \frac{AB}{20} = \frac{\sqrt{3}}{2} \\ \Rightarrow AB = \left(20\times \frac{\sqrt{3}}{2}\right) = 10\sqrt{3} \mathrm{cm} \\ \therefore \mathrm{BC} = 10 \mathrm{cm} \mathrm{and} \mathrm{AB} = 10\sqrt{3} \mathrm{cm} \end{gathered}$$

Question 30:

From the given right-angled triangle, we have:

$$\begin{gathered} \frac{BC}{AB}=\tan {30}^{\mathrm{o}} \\ \Rightarrow \frac{6}{AB}=\frac{1}{\sqrt{3}} \\ \Rightarrow AB=6\sqrt{3} \mathrm{cm} \\ \mathrm{Also}, \frac{BC}{AC}=\sin {30}^o \\ \Rightarrow \frac{6}{AC}=\frac{1}{2} \\ \Rightarrow AC=\left(2\times 6\right)=12 \mathrm{cm} \\ \therefore AB = 6\sqrt{3} \mathrm{cm} \mathrm{and} AC = 12 \mathrm{cm} \end{gathered}$$

Question 1:

From right-angled ∆ABC, we have:
​
$$\begin{gathered} \frac{BC}{AC}=\sin {45}^{\mathrm{o}} \\ \Rightarrow \frac{BC}{3\sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow BC=3 \mathrm{cm} \\ \mathrm{Also}, \frac{AB}{AC}=\cos {45}^o \\ \Rightarrow \frac{AB}{3\sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow AB=3 \mathrm{cm} \\ \therefore BC=3 \mathrm{cm} \mathrm{and} AB=3 \mathrm{cm} \end{gathered}$$

Question 2:

On substituting the values of various T-ratios, we get:
 sin 60^o cos 30^o + cos 60^o sin 30^o 
 $=\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$

Question 3:

Ans

Question 4:

On substituting the values of various T-ratios, we get:
 cos 60^o cos 30^o − sin 60^o sin 30^o 
 $=\left(\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$

Question 5:

On substituting the values of various T-ratios, we get:
 cos 45^o cos 30^o + sin 45^o  sin 30^o 
 $= \left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\times \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}\right) =\left(\frac{\sqrt{3} +1}{2\sqrt{2}}\right)$

Question 6:

As we know that,
$$\begin{gathered} \tan 30°=\frac{1}{\sqrt{3}} \\ \sec 30°=\frac{2}{\sqrt{3}} \\ \mathrm{cosec}60°=\frac{2}{\sqrt{3}} \\ \tan 60°=\sqrt{3} \\ \mathrm{By} \mathrm{substituting} \mathrm{these} \mathrm{values}, \mathrm{we} \mathrm{get} \\ \tan 30° \mathrm{cosec}60°+\tan 60° \sec 30°=\frac{1}{\sqrt{3}}\times \frac{2}{\sqrt{3}}+\sqrt{3}\times \frac{2}{\sqrt{3}} \\ =\frac{2}{3}+2 \\ =\frac{2+6}{3} \\ =\frac{8}{3} \\ \mathrm{Hence}, \tan 30° \mathrm{cosec}60°+\tan 60° \sec 30°=\frac{8}{3} \end{gathered}$$
 

Question 7:

On substituting the values of various T-ratios, we get:
 2 cos² 60^o + 3 sin² 45^o − 3 sin² 30^o + 2 cos² 90^o
 $$\begin{gathered} =2\times {\left(\frac{1}{2}\right)}^2+3\times {\left(\frac{1}{\sqrt{2}}\right)}^2-3\times {\left(\frac{1}{2}\right)}^2 + 2\times {\left(0\right)}^2 \\ =2\times \frac{1}{4}+3\times \frac{1}{2}-3\times \frac{1}{4}+0 \\ = \left(\frac{1}{2}+\frac{3}{2}-\frac{3}{4}\right)=\left(\frac{2+6-3}{4}\right)=\frac{5}{4} \end{gathered}$$

Question 8:

As we know that,
​$$\begin{gathered} \sin 30°=\frac{1}{2} \\ \cos 45°=\frac{1}{\sqrt{2}} \\ \tan 30°=\frac{1}{\sqrt{3}} \\ \sin 90°=1 \\ \cot 60°=\frac{1}{\sqrt{3}} \\ \mathrm{By} \mathrm{substituting} \mathrm{these} \mathrm{values}, \mathrm{we} \mathrm{get} \\ {\sin }^{2}30° {\cos }^{2}45°+4{\tan }^{2}30°+\frac{1}{2}{\sin }^{2}90°+\frac{1}{8}{\cot }^{2}60°={\left(\frac{1}{2}\right)}^2\times {\left(\frac{1}{\sqrt{2}}\right)}^2+4{\left(\frac{1}{\sqrt{3}}\right)}^2+\frac{1}{2}{\left(1\right)}^2+\frac{1}{8}{\left(\frac{1}{\sqrt{3}}\right)}^2 \\ =\left(\frac{1}{4}\right)\times \left(\frac{1}{2}\right)+4\left(\frac{1}{3}\right)+\frac{1}{2}\left(1\right)+\frac{1}{8}\left(\frac{1}{3}\right) \\ =\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24} \\ =\frac{3+32+12+1}{24} \\ =\frac{48}{24} \\ =2 \\ \mathrm{Hence}, {\sin }^{2}30° {\cos }^{2}45°+4{\tan }^{2}30°+\frac{1}{2}{\sin }^{2}90°+\frac{1}{8}{\cot }^{2}60°=2. \end{gathered}$$

Question 9:

On substituting the values of various T-ratios, we get:
  cot² 30^o − 2 cos² 30^o − $\frac{3}{4}$ sec² 45^o + $\frac{1}{4}$ cosec² 30^o
 $$\begin{gathered} = {\left(\sqrt{3}\right)}^2-2\times {\left(\frac{\sqrt{3}}{2}\right)}^2-\frac{3}{4}\times {\left(\sqrt{2}\right)}^2+\frac{1}{4}\times {\left(2\right)}^2 \\ = 3-2\times \frac{3}{4}-\frac{3}{4}\times 2+\frac{1}{4}\times 4 \\ = 3-\frac{3}{2}-\frac{3}{2}+1 \\ = 4-\left(\frac{3}{2}+\frac{3}{2}\right) \\ = 4-3 \\ = 1 \end{gathered}$$

Question 10:

As we know that,
​$$\begin{gathered} \cos 0°=1 \\ \cos 45°=\frac{1}{\sqrt{2}} \\ \sin 30°=\frac{1}{2} \\ \sin 90°=1 \\ \cos 60°=\frac{1}{2} \\ \mathrm{By} \mathrm{substituting} \mathrm{these} \mathrm{values}, \mathrm{we} \mathrm{get} \\ \left(\cos 0°+\cos 45°+\sin 30°\right)\left(\sin 90°-\cos 45°+\cos 60°\right)=\left(1+\frac{1}{\sqrt{2}}+\frac{1}{2}\right)\left(1-\frac{1}{\sqrt{2}}+\frac{1}{2}\right) \\ =\left(\left(1+\frac{1}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\right)\left(\left(1+\frac{1}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\right) \\ ={\left(1+\frac{1}{2}\right)}^2-{\left(\frac{1}{\sqrt{2}}\right)}^2 \\ =1+\frac{1}{4}+2\left(1\right)\left(\frac{1}{2}\right)-\frac{1}{2} \\ =1+\frac{1}{4}+1-\frac{1}{2} \\ =\frac{4+1+4-2}{4} \\ =\frac{7}{4} \\ \mathrm{Hence}, \left(\cos 0°+\cos 45°+\sin 30°\right)\left(\sin 90°-\cos 45°+\cos 60°\right)=\frac{7}{4}. \end{gathered}$$

Question 11:

(i)
 $$\begin{gathered} \mathrm{LHS}=\frac{1-\sin {60}^{\mathrm{o}}}{\cos {60}^{\mathrm{o}}}=\frac{1-{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}=\frac{\left({\displaystyle \frac{2-\sqrt{3}}{2}}\right)}{{\displaystyle \frac{1}{2}}}=\left({\displaystyle \frac{2-\sqrt{3}}{2}}\right)\times 2=2-\sqrt{3} \\ \mathrm{RHS}= \frac{\tan {60}^{\mathrm{o}}-1}{\tan {60}^{\mathrm{o}}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3} -1}{\sqrt{3} -1}=\frac{{\left(\sqrt{3}-1\right)}^2}{{\left(\sqrt{3}\right)}^2-1^2}=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3} \end{gathered}$$

Hence, LHS = RHS

∴ $\frac{1-\sin {60}^{\mathrm{o}}}{\cos {60}^{\mathrm{o}}}=\frac{\tan {60}^{\mathrm{o}}-1}{\tan {60}^{\mathrm{o}}+1}$

(ii)
 $$\begin{gathered} \mathrm{LHS} = \frac{\cos {30}^{\mathrm{o}}+\sin {60}^{\mathrm{o}}}{1+\sin {30}^{\mathrm{o}}+\cos {60}^{\mathrm{o}}}=\frac{\left({\displaystyle \frac{\sqrt{3}}{2}}+{\displaystyle \frac{\sqrt{3}}{2}} \right)}{\left(1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}} \right)}=\frac{{\displaystyle \frac{\sqrt{3}+\sqrt{3}}{2}}}{{\displaystyle \frac{2+1+1}{2}}}=\frac{\sqrt{3}}{2} \\ \mathrm{Also}, \mathrm{RHS} = \cos {30}^{\mathrm{o}}=\frac{\sqrt{3}}{2} \end{gathered}$$

Hence, LHS = RHS

 ∴  $\frac{\cos {30}^{\mathrm{o}}+\sin {60}^{\mathrm{o}}}{1+\sin {30}^{\mathrm{o}}+\cos {60}^{\mathrm{o}}}=\cos {30}^{\mathrm{o}}$1−sin60°cos60°

Question 12:

(i) sin 60^o cos 30^o − cos 60^o sin 30^o
 $$\begin{gathered} =\left(\frac{\sqrt{3}}{2}\right)\times \left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2} \\ \mathrm{Also}, \sin {30}^{\mathrm{o}}=\frac{1}{2} \end{gathered}$$
∴ sin 60^o cos 30^o − cos 60^o sin 30^o = sin 30^o

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

​$$\begin{gathered} =\left(\frac{1}{2}\right)\times \left(\frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\times \left(\frac{1}{2}\right)=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2} \\ \mathrm{Also}, \cos {30}^{\mathrm{o}} =\frac{\sqrt{3}}{2} \end{gathered}$$
∴ cos 60^o cos 30^o + sin 60^o sin 30^o = cos 30^o

(iii) 2 sin 30^o cos 30^o
$$\begin{gathered} =2\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \\ \mathrm{Also}, \sin {60}^{\mathrm{o}} =\frac{\sqrt{3}}{2} \end{gathered}$$
∴ 2 sin 30^o cos 30^o = sin 60^o

(iv) 2 sin 45^o cos 45^o
$=2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=1$
Also, sin 90^o = 1
∴ 2 sin 45^o cos 45^o = sin 90^o

Question 13:

A = 45^o
⇒ 2A = 2 $\times$ 45^o = 90^o

(i) sin 2A = sin 90^o = 1
2 sin A cos A = 2 sin 45^o cos 45^o = $2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}} = 1$
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90^o = 0
2 cos²A − 1 = 2 cos² 45^o − 1 =  $2\times {\left(\frac{1}{\sqrt{2}}\right)}^2 - 1 = 2\times \frac{1}{2} -1 = 1-1 = 0$
Now, 1 − 2 sin²A = $1-2\times {\left(\frac{1}{\sqrt{2}}\right)}^2 = 1 - 2\times \frac{1}{2} =1 - 1 = 0$
∴ cos 2A =  2 cos²A − 1 = 1 − 2 sin²A

Question 14:

A = 30^o
⇒ 2A = 2 $\times$ 30^o = 60^o

​(i) sin 2A = sin 60^o = $\frac{\sqrt{3}}{2}$
$\frac{2 \tan A}{1+{\tan }^2 A}=\frac{2 \tan {30}^o}{1+{\tan }^2 {30}^o}=\frac{2\times \left({\displaystyle \frac{1}{\sqrt{3}}}\right)}{1+{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2 }=\frac{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}{1 + {\displaystyle \frac{1}{3}}}=\frac{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}{{\displaystyle \frac{4}{3}}}=\left({\displaystyle \frac{2}{\sqrt{3}}}\right)\times \frac{3}{4}=\frac{\sqrt{3}}{2}$
∴ $\sin 2A=\frac{2\tan A}{1+{\tan }^{2}A}$

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

Question 15:

A = 60^o and B = 30^o
Now, A + B = 60^o + 30^o​ = 90^o
Also, A − B = 60^o − 30^o = 30^o

(i) sin (A + B) = sin 90^o = 1
sin A cos B + cos A sin B = sin 60^o cos 30^o + cos 60^o sin 30^o
= $\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{2}\right)=\left(\frac{3}{4} + \frac{1}{4}\right)=1$
∴ sin (A + B) = sin A cos B + cos A sin B


(ii) cos (A + B) = cos 90^o = 0
cos A cos B − sin A sin B =cos 60^o cos 30^o − sin 60^o sin 30^o $=\left(\frac{1}{2}\times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\times \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}\right) = 0$
∴​ cos (A + B) = cos A cos B − sin A sin B
 

Question 16:

(i) sin (A − B) = sin 30^o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60^o cos 30^o − cos 60^o sin 30^o
= $\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}- \frac{1}{2}\times \frac{1}{2}\right) = \left(\frac{3}{4} - \frac{1}{4}\right) = \frac{2}{4} = \frac{1}{2}$
∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (A − B) = cos 30^o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60^o cos 30^o + sin 60^o sin 30^o
= $\left(\frac{1}{2}\times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\times \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}\right) = 2\times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$
∴​ cos (A − B) = cos A cos B + sin A sin B

(iii) tan (A − B) = tan 30^o = $\frac{1}{\sqrt{3}}$
 $\frac{\tan A-\tan B}{1+\tan A \tan B}=\frac{\tan {60}^o- \tan {30}^o}{1 +\tan {60}^o \tan {30}^o}= \frac{\sqrt{3} -\frac{1}{\sqrt{3}} }{1+\left(\sqrt{3}\times \frac{1}{\sqrt{3}}\right)}=\frac{1}{2}\times \left(\frac{3- 1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}$
∴​ tan (A − B) = $\frac{\tan A-\tan B}{1+\tan A \tan B}$