RS Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 10 - Trignometric Ratios

Question 1:

Question 2:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that sin $\theta$ = $\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$= $\frac{AB}{AC}$ = $\frac{\sqrt{3}}{2 }$.

So, if AB = $\sqrt{3}k$, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² 
⇒ BC² = AC²$-$ AB² = (2k)² $-$ ($\sqrt{3}k$)²
⇒ BC² = 4k²$-$ 3k² = k²
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
   cos $\theta$  = $\frac{BC}{AC}$ = $\frac{k}{2k} = \frac{1}{2}$
   tan $\theta$  = $\frac{AB}{BC} = \frac{\sqrt{3}k}{k} = \sqrt{3}$

 ∴ cot $\theta$  = $\frac{1}{\tan \theta } = \frac{1}{\sqrt{3}}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \frac{2}{\sqrt{3}}$ and sec $\theta$  = $\frac{1}{\cos \theta } = 2$

Question 3:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cos $\theta$ = $\frac{\mathrm{Base}}{\mathrm{hypotenuse}}$= $\frac{BC}{AC}$ = $\frac{7}{25}$.

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² 
⇒ AB² = AC²$-$ BC² = (25k)² $-$ (7k)².
⇒ AB² = 625k²$-$ 49k² = 576k²
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
   sin $\theta$ = $\frac{AB}{AC}$ = $\frac{24k}{25k} = \frac{24}{25}$
   tan $\theta$ = $\frac{AB}{BC} = \frac{24k}{7k} = \frac{24}{7}$
 ∴ cot $\theta$ = $\frac{1}{\tan \theta } = \frac{7}{24}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \frac{25}{24}$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{25}{7}$

Question 4:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Base}}$ = $\frac{AB}{BC}$ = $\frac{15}{8}$.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (15k)² + (8k)²
⇒ AC² = 225k² + 64k² = 289k²
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
  sin $\theta$  = $\frac{AB}{AC}$ = $\frac{15k}{17k} = \frac{15}{17}$
  cos $\theta$  = $\frac{BC}{AC} = \frac{8k}{17k} = \frac{8}{17}$

∴ cot $\theta$  = $\frac{1}{\tan \theta } = \frac{8}{15}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \frac{17}{15}$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{17}{8}$

Question 5:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cot $\theta$ = $\frac{\mathrm{base}}{\mathrm{Perpendicular}}$ = $\frac{BC}{AB}$ = 2.


So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC² = AB² + BC² = (2k)² + (k)²
⇒ AC² = 4k² + k² = 5k²
⇒ AC = $\sqrt{5}$k
Now, finding the other T-ratios using their definitions, we get:
   sin $\theta$  = $\frac{AB}{AC}$ = $\frac{k}{\sqrt{5}k} = \frac{1}{\sqrt{5}}$
   cos $\theta$  = $\frac{BC}{AC} = \frac{2k}{\sqrt{5}k} = \frac{2}{\sqrt{5}}$

∴ tan $\theta$  = $\frac{1}{cot \theta } = \frac{1}{2}$, cosec $\theta$ = $\frac{1}{\sin \theta } = \sqrt{5}$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{\sqrt{5}}{2}$

Question 6:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cosec $\theta$ = $\frac{\mathrm{Hypotenuse}}{\mathrm{Perpendicular}}$ = $\frac{AC}{AB}$= $\frac{\sqrt{10}}{1}$.

So, if AC = ($\sqrt{10}$)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC² = AB² + BC² 
⇒ BC² = AC²$-$ AB² = 10k²$-$k²
⇒ BC² = 9k²
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
   tan $\theta$  = $\frac{AB}{BC}$ = $\frac{k}{3k} = \frac{1}{3}$

   cos $\theta$  = $\frac{BC}{AC} = \frac{3k}{\sqrt{10}k} = \frac{3}{\sqrt{10}}$

 ∴ $\sin \theta =\frac{1}{\cos ec \theta }=\frac{1}{\sqrt{10}}$, cot $\theta$  = $\frac{1}{\tan \theta } = 3$ and sec $\theta$  = $\frac{1}{\cos \theta } = \frac{\sqrt{10}}{3}$

Question 7:

We have $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$,

As,

$$\begin{gathered} {\cos }^2\theta =1-{\sin }^2\theta \\ =1-{\left(\frac{a^2-b^2}{a^2+b^2}\right)}^2 \\ =\frac{1}{1}-\frac{{\left(a^2-b^2\right)}^2}{{\left(a^2+b^2\right)}^2} \\ =\frac{{\left(a^2+b^2\right)}^2-{\left(a^2-b^2\right)}^2}{{\left(a^2+b^2\right)}^2} \\ =\frac{\left[\left(a^2+b^2\right)-\left(a^2-b^2\right)\right]\left[\left(a^2+b^2\right)+\left(a^2-b^2\right)\right]}{{\left(a^2+b^2\right)}^2} \end{gathered}$$
$$\begin{gathered} =\frac{\left[a^2+b^2-a^2+b^2\right]\left[a^2+b^2+a^2-b^2\right]}{{\left(a^2+b^2\right)}^2} \\ =\frac{\left[2b^2\right]\left[2a^2\right]}{{\left(a^2+b^2\right)}^2} \\ \Rightarrow {\cos }^2\theta =\frac{4a^2{b}^2}{{\left(a^2+b^2\right)}^2} \\ \Rightarrow \cos \theta =\sqrt{\frac{4a^2{b}^2}{{\left(a^2+b^2\right)}^2}} \\ \Rightarrow \cos \theta =\frac{2ab}{\left(a^2+b^2\right)} \end{gathered}$$

Also,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ =\frac{\left({\displaystyle \frac{a^2-b^2}{a^2+b^2}}\right)}{\left({\displaystyle \frac{2ab}{a^2+b^2}}\right)} \\ =\frac{a^2-b^2}{2ab} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{cosec}\theta =\frac{1}{\sin \theta } \\ =\frac{1}{\left({\displaystyle \frac{a^2-b^2}{a^2+b^2}}\right)} \\ =\frac{a^2+b^2}{a^2-b^2} \end{gathered}$$

Also,

$$\begin{gathered} \sec \theta =\frac{1}{\cos \theta } \\ =\frac{1}{\left({\displaystyle \frac{2ab}{a^2+b^2}}\right)} \\ =\frac{a^2+b^2}{2ab} \end{gathered}$$

And,

$$\begin{gathered} \cot \theta =\frac{1}{\tan \theta } \\ =\frac{1}{\left({\displaystyle \frac{a^2-b^2}{2ab}}\right)} \\ =\frac{2ab}{a^2-b^2} \end{gathered}$$

Question 8:

$$\begin{gathered} \mathrm{Given}: \sin \theta =\frac{c}{\sqrt{c^2+d^2}} \\ \mathrm{Since}, \sin \theta =\frac{P}{H} \\ \Rightarrow P=c \mathrm{and} H=\sqrt{c^2+d^2} \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow c^2+B^2=c^2+d^2 \\ \Rightarrow B^2=d^2 \\ \Rightarrow B=d \\ \mathrm{Therefore}, \\ \cos \theta =\frac{B}{H}=\frac{d}{\sqrt{c^2+d^2}} \\ \tan \theta =\frac{P}{B}=\frac{c}{d} \\ \mathrm{Hence}, \cos \theta =\frac{d}{\sqrt{c^2+d^2}} \mathrm{and} \tan \theta =\frac{c}{d}. \end{gathered}$$

Question 9:

$$\begin{gathered} \mathrm{Given}: \sqrt{3}\tan \theta =1 \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \\ \mathrm{Since}, \tan \theta =\frac{P}{B} \\ \Rightarrow P=1 \mathrm{and} B=\sqrt{3} \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 1^2+{\left(\sqrt{3}\right)}^2=H^2 \\ \Rightarrow H^2=1+3=4 \\ \Rightarrow H=2 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{1}{2} \\ \cos \theta =\frac{B}{H}=\frac{\sqrt{3}}{2} \\ {\cos }^2\theta -{\sin }^2\theta ={\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2 \\ =\frac{3}{4}-\frac{1}{4}=\frac{2}{4} \\ =\frac{1}{2} \\ \mathrm{Hence}, {\cos }^2\theta -{\sin }^2\theta =\frac{1}{2}. \end{gathered}$$

Question 10:

$$\begin{gathered} \mathrm{Given}: 4\tan \theta =3 \\ \Rightarrow \tan \theta =\frac{3}{4} \\ \mathrm{Since}, \tan \theta =\frac{P}{B} \\ \Rightarrow P=3 \mathrm{and} B=4 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 3^2+4^2=H^2 \\ \Rightarrow H^2=9+16=25 \\ \Rightarrow H=5 \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{3}{5} \\ \cos \theta =\frac{B}{H}=\frac{4}{5} \\ \sin \theta \times \cos \theta =\frac{3}{5}\times \frac{4}{5} \\ =\frac{12}{25} \\ \mathrm{Hence}, \sin \theta \times \cos \theta =\frac{12}{25}. \end{gathered}$$

Question 11:

$$\begin{gathered} \mathrm{LHS}=\left(\sec \theta +\tan \theta \right) \\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\sqrt{1-{\sin }^2\theta }} \\ =\frac{\left(1+{\displaystyle \frac{a}{b}}\right)}{\sqrt{1-{\left({\displaystyle \frac{a}{b}}\right)}^2}} \end{gathered}$$
$$\begin{gathered} =\frac{\left({\displaystyle \frac{1}{1}+\frac{a}{b}}\right)}{\sqrt{{\displaystyle \frac{1}{1}-\frac{a^2}{b^2}}}} \\ =\frac{\left({\displaystyle \frac{b+a}{b}}\right)}{\sqrt{{\displaystyle \frac{b^2-a^2}{b^2}}}} \\ =\frac{\left({\displaystyle \frac{b+a}{b}}\right)}{\left({\displaystyle \frac{\sqrt{b^2-a^2}}{b}}\right)} \\ =\frac{\left({\displaystyle b+a}\right)}{{\displaystyle \sqrt{\left(b+a\right)\left(b-a\right)}}} \end{gathered}$$
$$\begin{gathered} =\frac{\left({\displaystyle b+a}\right)}{{\displaystyle \sqrt{\left(b+a\right)}\sqrt{\left(b-a\right)}}} \\ =\frac{\sqrt{\left(b+a\right)}}{{\displaystyle \sqrt{\left(b-a\right)}}} \\ =\sqrt{\frac{b+a}{b-a}} \\ =\mathrm{RHS} \end{gathered}$$

Question 12:

It is given that tan $\theta = \frac{a}{b}$.

LHS = $\frac{a \sin \theta - b \cos \theta }{a \sin \theta + b \cos \theta }$
 Dividing the numerator and denominator by cos $\theta$, we get:

 $\frac{a \tan \theta - b}{a \tan \theta + b}$       (∵ tan $\theta = \frac{\sin \theta }{\cos \theta }$)
Now, substituting the value of tan $\theta$ in the above expression, we get:
 $$\begin{gathered} \frac{a\left({\displaystyle \frac{a}{b}}\right) - b}{a\left({\displaystyle \frac{a}{b}}\right) + b} \\ = \frac{{\displaystyle \frac{a^2}{b}} - b}{{\displaystyle \frac{a^2}{b}} + b} \\ = \frac{a^2 - b^2}{a^2 + b^2} = \mathrm{RHS} \end{gathered}$$
  i.e., LHS = RHS

 Hence proved.

Question 13:

$$\begin{gathered} \mathrm{Given}: \sin \theta =\frac{12}{13} \\ \mathrm{Since}, \sin \theta =\frac{P}{H} \\ \Rightarrow P=12 \mathrm{and} H=13 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow {12}^2+B^2={13}^2 \\ \Rightarrow B^2=169-144 \\ \Rightarrow B^2=25 \\ \Rightarrow B=5 \\ \mathrm{Therefore}, \\ \cos \theta =\frac{B}{H}=\frac{5}{13} \\ \mathrm{Now}, \\ \left(\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }\right)=\left(\frac{2\left({\displaystyle \frac{12}{13}}\right)-3\left({\displaystyle \frac{5}{13}}\right)}{4\left({\displaystyle \frac{12}{13}}\right)-9\left({\displaystyle \frac{5}{13}}\right)}\right) \\ =\left(\frac{{\displaystyle \frac{24}{13}}-{\displaystyle \frac{15}{13}}}{{\displaystyle \frac{48}{13}}-{\displaystyle \frac{45}{13}}}\right) \\ =\left(\frac{{\displaystyle 24-15}}{{\displaystyle 48-45}}\right) \\ =\frac{{\displaystyle 9}}{{\displaystyle 3}} \\ =3 \\ \mathrm{Hence}, \left(\frac{2\sin \theta -3\cos \theta }{4\sin \theta -9\cos \theta }\right)=3. \end{gathered}$$

Question 14:

$$\begin{gathered} \mathrm{Given}: \tan \theta =\frac{1}{2} \\ \mathrm{Since}, \tan \theta =\frac{P}{B} \\ \Rightarrow P=1 \mathrm{and} B=2 \\ \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 1^2+2^2=H^2 \\ \Rightarrow H^2=1+4 \\ \Rightarrow H^2=5 \\ \Rightarrow H=\sqrt{5} \\ \mathrm{Therefore}, \\ \sin \theta =\frac{P}{H}=\frac{1}{\sqrt{5}} \\ \cos \theta =\frac{B}{H}=\frac{2}{\sqrt{5}} \\ \mathrm{Now}, \\ \left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{1+\cos \theta }\right)=\left(\frac{{\displaystyle \frac{2}{\sqrt{5}}}}{{\displaystyle \frac{1}{\sqrt{5}}}}+\frac{{\displaystyle \frac{1}{\sqrt{5}}}}{1+{\displaystyle \frac{2}{\sqrt{5}}}}\right) \\ =\left(\frac{2}{1}+\frac{{\displaystyle \frac{1}{\sqrt{5}}}}{{\displaystyle \frac{\sqrt{5}+2}{\sqrt{5}}}}\right) \\ =\left(\frac{2}{1}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{5}+2}}\right) \\ =\left(2+\left(\frac{{\displaystyle 1}}{{\displaystyle \sqrt{5}+2}}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}\right)\right) \\ =\left(2+\left(\frac{\sqrt{5}-2}{5-4}\right)\right) \\ =\left(2+\sqrt{5}-2\right) \\ =\sqrt{5} \\ \mathrm{Hence}, \left(\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{1+\cos \theta }\right)=\sqrt{5}. \end{gathered}$$

Question 15:

$$\begin{gathered} \mathrm{LHS}=\left(3\cos \alpha -4{\cos }^3\alpha \right) \\ =\cos \alpha \left(3-4{\cos }^2\alpha \right) \\ =\sqrt{1-{\sin }^2\alpha }\left[3-4\left(1-{\sin }^2\alpha \right)\right] \\ =\sqrt{1-{\left(\frac{1}{2}\right)}^2}\left[3-4\left(1-{\left(\frac{1}{2}\right)}^2\right)\right] \\ =\sqrt{\frac{1}{1}-\frac{1}{4}}\left[3-4\left(\frac{1}{1}-\frac{1}{4}\right)\right] \\ =\sqrt{\frac{3}{4}}\left[3-4\left(\frac{3}{4}\right)\right] \\ =\sqrt{\frac{3}{4}}\left[3-3\right] \\ =\sqrt{\frac{3}{4}}\left[0\right] \\ =0 \\ =\mathrm{RHS} \end{gathered}$$

Question 16:

It is given that cot $\theta = \frac{2}{3}$.

LHS  = $\frac{4 \sin \theta - 3 \cos \theta }{2 \sin \theta + 6 \cos \theta }$
Dividing the above expression by sin $\theta$, we get:
$\frac{4 - 3 cot \theta }{2 + 6 cot \theta }$                     [∵ cot $\theta = \frac{\cos \theta }{\sin \theta }$]
Now, substituting the values of cot $\theta$ in the above expression, we get:
$$\begin{gathered} \frac{4 - 3\left({\displaystyle \frac{2}{3}}\right)}{2 + 6\left({\displaystyle \frac{2}{3}}\right)} \\ = \frac{4 - 2}{2 + 4} = \frac{2}{6}=\frac{1}{3} \end{gathered}$$
 i.e., LHS = RHS
 
Hence proved.

Question 17:

It is given that sec $\theta$ = $\frac{17}{8}$.

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that cos $\theta$ = $\frac{1}{sec \theta }= \frac{8}{17} = \frac{BC}{AC}$
 
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC² = AB² + BC²
⇒ AB² = AC²$-$ BC² = (17k)²$-$ (8k)²
⇒ AB² = 289k²$-$ 64k² = 225k²
⇒ AB = 15k.

Now, tan $\theta$  = $\frac{AB}{BC} = \frac{15}{8}$ and sin $\theta$ = $\frac{AB}{AC} = \frac{15k}{17k}= \frac{15}{17}$

The given expression is $\frac{3 - 4{\sin }^2\theta }{4{\cos }^2\theta - 3} = \frac{3 - {\tan }^2\theta }{1 - 3{\tan }^2\theta }$.
 
 Substituting the values in the above expression, we get:
 $$\begin{gathered} \mathrm{LHS}= \frac{3 - 4{\left({\displaystyle \frac{15}{17}}\right)}^2}{4{\left({\displaystyle \frac{8}{17}}\right)}^2 - 3} \\ = \frac{3 - {\displaystyle \frac{900}{289}}}{{\displaystyle \frac{256}{289}}- 3} \\ = \frac{867-900}{256-867}= \frac{-33}{-611}=\frac{33}{611} \end{gathered}$$

$$\begin{gathered} \mathrm{RHS} = \frac{3-{\left({\displaystyle \frac{15}{8}}\right)}^2}{1-3{\left({\displaystyle \frac{15}{8}}\right)}^2} \\ =\frac{3-{\displaystyle \frac{225}{64}}}{1-{\displaystyle \frac{675}{64}}} \\ =\frac{192-225}{64-675}=\frac{-33}{-611}=\frac{33}{611} \end{gathered}$$

∴ LHS = RHS
Hence proved.

Question 18:

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{AB}{BC}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
 AC² = AB² + BC²
⇒ AC²= (20k)² + (21k)²
⇒ AC² = 841k²
⇒  AC = 29k
Now, sin $\theta$ = $\frac{AB}{AC} = \frac{20}{29}$ and cos $\theta$ = $\frac{BC}{AC} = \frac{21}{29}$

Substituting these values in the given expression, we get:
 $$\begin{gathered} \mathrm{LHS}=\frac{1 - \sin \theta + \cos \theta }{1 + \sin \theta + \cos \theta } \\ = \frac{1 - {\displaystyle \frac{20}{29}} + {\displaystyle \frac{21}{29}}}{1 + {\displaystyle \frac{20}{29}} + {\displaystyle \frac{21}{29}}} \\ = \frac{{\displaystyle \frac{29 - 20 + 21}{29}}}{{\displaystyle \frac{29 + 20 + 21}{29}}}= \frac{30}{70} = \frac{3}{7} = \mathrm{RHS} \end{gathered}$$
∴ LHS = RHS

Hence proved.