Exercise 6.3 — Class 9 Mathematics Solutions

Solution

Solution

1. Turn the given ratio into sides.

   Since $\sin \theta ={\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}}$, take a right-angled triangle with angle $\theta$ whose perpendicular is $2$ and hypotenuse is $3$. Let $P$, $B$ and $H$ stand for the perpendicular, base and hypotenuse. $P=2,H=3$

2. Find the third side.

   By Pythagoras theorem, $H^2=P^2+B^2$; a length is positive, so take the positive root.

   $$\begin{array}{cc}{B}^2& =H^2-P^2\\ & =3^2-2^2\\ & =9-4\\ & =5\\ ⟹B& =\sqrt{5}\end{array}$$

3. Write the remaining ratios.

   $\theta$ lies in the first quadrant, so every ratio is positive. Use the definitions and rationalise the denominators.

   $$\begin{array}{cc}\cos \theta & =\frac{B}{H}\\ & =\frac{\sqrt{5}}{3}\end{array}$$ $$\begin{array}{cc}\tan \theta & =\frac{P}{B}\\ & =\frac{2}{\sqrt{5}}\\ & =\frac{2\sqrt{5}}{5}\end{array}$$ $$\begin{array}{cc}\text{cosec}\theta & =\frac{H}{P}\\ & =\frac{3}{2}\end{array}$$ $$\begin{array}{cc}\sec \theta & =\frac{H}{B}\\ & =\frac{3}{\sqrt{5}}\\ & =\frac{3\sqrt{5}}{5}\end{array}$$ $$\begin{array}{cc}\cot \theta & =\frac{B}{P}\\ & =\frac{\sqrt{5}}{2}\end{array}$$

Solution

1. Turn the given ratio into sides.

   Since $\cos \theta ={\displaystyle \frac{\text{base}}{\text{hypotenuse}}}$, take a right-angled triangle with angle $\theta$ whose base is $3$ and hypotenuse is $4$. Let $P$, $B$ and $H$ stand for the perpendicular, base and hypotenuse. $B=3,H=4$

2. Find the third side.

   By Pythagoras theorem, $H^2=P^2+B^2$; a length is positive, so take the positive root.

   $$\begin{array}{cc}{P}^2& =H^2-B^2\\ & =4^2-3^2\\ & =16-9\\ & =7\\ ⟹P& =\sqrt{7}\end{array}$$

3. Write the remaining ratios.

   $\theta$ lies in the first quadrant, so every ratio is positive. Use the definitions and rationalise the denominators.

   $$\begin{array}{cc}\sin \theta & =\frac{P}{H}\\ & =\frac{\sqrt{7}}{4}\end{array}$$ $$\begin{array}{cc}\tan \theta & =\frac{P}{B}\\ & =\frac{\sqrt{7}}{3}\end{array}$$ $$\begin{array}{cc}\text{cosec}\theta & =\frac{H}{P}\\ & =\frac{4}{\sqrt{7}}\\ & =\frac{4\sqrt{7}}{7}\end{array}$$ $$\begin{array}{cc}\sec \theta & =\frac{H}{B}\\ & =\frac{4}{3}\end{array}$$ $$\begin{array}{cc}\cot \theta & =\frac{B}{P}\\ & =\frac{3}{\sqrt{7}}\\ & =\frac{3\sqrt{7}}{7}\end{array}$$

Solution

1. Turn the given ratio into sides.

   Since $\tan \theta ={\displaystyle \frac{\text{perpendicular}}{\text{base}}}$, take a right-angled triangle with angle $\theta$ whose perpendicular is $1$ and base is $2$. Let $P$, $B$ and $H$ stand for the perpendicular, base and hypotenuse. $P=1,B=2$

2. Find the third side.

   By Pythagoras theorem, $H^2=P^2+B^2$; a length is positive, so take the positive root.

   $$\begin{array}{cc}{H}^2& =P^2+B^2\\ & =1^2+2^2\\ & =1+4\\ & =5\\ ⟹H& =\sqrt{5}\end{array}$$

3. Write the remaining ratios.

   $\theta$ lies in the first quadrant, so every ratio is positive. Use the definitions and rationalise the denominators.

   $$\begin{array}{cc}\sin \theta & =\frac{P}{H}\\ & =\frac{1}{\sqrt{5}}\\ & =\frac{\sqrt{5}}{5}\end{array}$$ $$\begin{array}{cc}\cos \theta & =\frac{B}{H}\\ & =\frac{2}{\sqrt{5}}\\ & =\frac{2\sqrt{5}}{5}\end{array}$$ $$\begin{array}{cc}\text{cosec}\theta & =\frac{H}{P}\\ & =\sqrt{5}\end{array}$$ $$\begin{array}{cc}\sec \theta & =\frac{H}{B}\\ & =\frac{\sqrt{5}}{2}\end{array}$$ $$\begin{array}{cc}\cot \theta & =\frac{B}{P}\\ & =2\end{array}$$

Solution

1. Turn the given ratio into sides.

   Write $\sec \theta ={\displaystyle \frac{3}{1}}$. Since $\sec \theta ={\displaystyle \frac{\text{hypotenuse}}{\text{base}}}$, take a right-angled triangle with angle $\theta$ whose hypotenuse is $3$ and base is $1$. Let $P$, $B$ and $H$ stand for the perpendicular, base and hypotenuse. $H=3,B=1$

2. Find the third side.

   By Pythagoras theorem, $H^2=P^2+B^2$; a length is positive, so take the positive root.

   $$\begin{array}{cc}{P}^2& =H^2-B^2\\ & =3^2-1^2\\ & =9-1\\ & =8\\ ⟹P& =\sqrt{8}\\ & =2\sqrt{2}\end{array}$$

3. Write the remaining ratios.

   $\theta$ lies in the first quadrant, so every ratio is positive. Use the definitions and rationalise the denominators.

   $$\begin{array}{cc}\sin \theta & =\frac{P}{H}\\ & =\frac{2\sqrt{2}}{3}\end{array}$$ $$\begin{array}{cc}\cos \theta & =\frac{B}{H}\\ & =\frac{1}{3}\end{array}$$ $$\begin{array}{cc}\tan \theta & =\frac{P}{B}\\ & =2\sqrt{2}\end{array}$$ $$\begin{array}{cc}\text{cosec}\theta & =\frac{H}{P}\\ & =\frac{3}{2\sqrt{2}}\\ & =\frac{3\sqrt{2}}{4}\end{array}$$ $$\begin{array}{cc}\cot \theta & =\frac{B}{P}\\ & =\frac{1}{2\sqrt{2}}\\ & =\frac{\sqrt{2}}{4}\end{array}$$

Solution

1. Turn the given ratio into sides.

   Write $\cot \theta =\sqrt{{\displaystyle \frac{3}{2}}}={\displaystyle \frac{\sqrt{3}}{\sqrt{2}}}$. Since $\cot \theta ={\displaystyle \frac{\text{base}}{\text{perpendicular}}}$, take a right-angled triangle with angle $\theta$ whose base is $\sqrt{3}$ and perpendicular is $\sqrt{2}$. Let $P$, $B$ and $H$ stand for the perpendicular, base and hypotenuse. $B=\sqrt{3},P=\sqrt{2}$

2. Find the third side.

   By Pythagoras theorem, $H^2=P^2+B^2$; a length is positive, so take the positive root.

   $$\begin{array}{cc}{H}^2& =P^2+B^2\\ & =(\sqrt{2}{)}^2+(\sqrt{3}{)}^2\\ & =2+3\\ & =5\\ ⟹H& =\sqrt{5}\end{array}$$

3. Write the remaining ratios.

   $\theta$ lies in the first quadrant, so every ratio is positive. Use the definitions and rationalise the denominators.

   $$\begin{array}{cc}\sin \theta & =\frac{P}{H}\\ & =\frac{\sqrt{2}}{\sqrt{5}}\\ & =\frac{\sqrt{10}}{5}\end{array}$$ $$\begin{array}{cc}\cos \theta & =\frac{B}{H}\\ & =\frac{\sqrt{3}}{\sqrt{5}}\\ & =\frac{\sqrt{15}}{5}\end{array}$$ $$\begin{array}{cc}\tan \theta & =\frac{P}{B}\\ & =\frac{\sqrt{2}}{\sqrt{3}}\\ & =\frac{\sqrt{6}}{3}\end{array}$$ $$\begin{array}{cc}\text{cosec}\theta & =\frac{H}{P}\\ & =\frac{\sqrt{5}}{\sqrt{2}}\\ & =\frac{\sqrt{10}}{2}\end{array}$$ $$\begin{array}{cc}\sec \theta & =\frac{H}{B}\\ & =\frac{\sqrt{5}}{\sqrt{3}}\\ & =\frac{\sqrt{15}}{3}\end{array}$$

Solution

1. Expand the square.

   Use $(a+b{)}^2=a^2+2ab+b^2$ on the L.H.S.

   $$\begin{array}{cc}\text{L.H.S}& =(\sin \theta +\cos \theta {)}^2\\ & ={\sin }^2\theta +2\sin \theta \cos \theta +{\cos }^2\theta \\ & =({\sin }^2\theta +{\cos }^2\theta )+2\sin \theta \cos \theta \end{array}$$

2. Apply the Pythagorean identity.

   Replace ${\sin }^2\theta +{\cos }^2\theta$ by $1$, since ${\sin }^2\theta +{\cos }^2\theta =1$.

   $$\begin{array}{cc}\text{L.H.S}& =1+2\sin \theta \cos \theta \\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Recognise the quotient.

   By definition, $\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}$.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{\cos \theta }{\sin \theta }\\ & =\cot \theta \end{array}$$

2. Use the reciprocal relation.

   $\cot \theta$ is the reciprocal of $\tan \theta$.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{1}{\tan \theta }\\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Replace the reciprocal ratios.

   Since $\text{cosec}\theta ={\displaystyle \frac{1}{\sin \theta }}$ and $\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$, dividing by them is the same as multiplying by $\sin \theta$ and $\cos \theta$.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{\sin \theta }{\text{cosec}\theta }+\frac{\cos \theta }{\sec \theta }\\ & =\sin \theta \cdot \sin \theta +\cos \theta \cdot \cos \theta \\ & ={\sin }^2\theta +{\cos }^2\theta \end{array}$$

2. Apply the Pythagorean identity.

   Use ${\sin }^2\theta +{\cos }^2\theta =1$.

   $$\begin{array}{cc}\text{L.H.S}& =1\\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Write sine in terms of cosine.

   From ${\sin }^2\theta +{\cos }^2\theta =1$, ${\sin }^2\theta =1-{\cos }^2\theta$.

   $$\begin{array}{cc}\text{L.H.S}& ={\cos }^2\theta -{\sin }^2\theta \\ & ={\cos }^2\theta -(1-{\cos }^2\theta )\end{array}$$

2. Simplify.

   Remove the bracket and collect like terms.

   $$\begin{array}{cc}\text{L.H.S}& ={\cos }^2\theta -1+{\cos }^2\theta \\ & =2{\cos }^2\theta -1\\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Write cosine in terms of sine.

   From ${\sin }^2\theta +{\cos }^2\theta =1$, ${\cos }^2\theta =1-{\sin }^2\theta$.

   $$\begin{array}{cc}\text{L.H.S}& ={\cos }^2\theta -{\sin }^2\theta \\ & =(1-{\sin }^2\theta )-{\sin }^2\theta \end{array}$$

2. Simplify.

   Collect the ${\sin }^2\theta$ terms.

   $$\begin{array}{cc}\text{L.H.S}& =1-2{\sin }^2\theta \\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Multiply top and bottom by the conjugate.

   Multiplying by ${\displaystyle \frac{1+\sin \theta }{1+\sin \theta }}$ (which equals 1) lets us use $(a-b)(a+b)=a^2-b^2$ in the numerator.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{1-\sin \theta }{\cos \theta }\times \frac{1+\sin \theta }{1+\sin \theta }\\ & =\frac{1-{\sin }^2\theta }{\cos \theta (1+\sin \theta )}\end{array}$$

2. Apply the Pythagorean identity.

   From ${\sin }^2\theta +{\cos }^2\theta =1$, $1-{\sin }^2\theta ={\cos }^2\theta$.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{{\cos }^2\theta }{\cos \theta (1+\sin \theta )}\\ & =\frac{\cos \theta }{1+\sin \theta }\\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Change to sine and cosine.

   Use $\sec \theta ={\displaystyle \frac{1}{\cos \theta }}$ and $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$; the two fractions share a denominator.

   $$\begin{array}{cc}\text{L.H.S}& =(\sec \theta -\tan \theta {)}^2\\ & ={(\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta })}^2\\ & =\frac{(1-\sin \theta {)}^2}{{\cos }^2\theta }\end{array}$$

2. Rewrite the denominator.

   From ${\sin }^2\theta +{\cos }^2\theta =1$, ${\cos }^2\theta =1-{\sin }^2\theta$, which factorises as a difference of squares.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{(1-\sin \theta {)}^2}{1-{\sin }^2\theta }\\ & =\frac{(1-\sin \theta {)}^2}{(1-\sin \theta )(1+\sin \theta )}\end{array}$$

3. Cancel the common factor.

   Cancel one factor of $1-\sin \theta$.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{1-\sin \theta }{1+\sin \theta }\\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Change to sine and cosine.

   Use $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$ and $\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}$, then add over the common denominator $\sin \theta \cos \theta$.

   $$\begin{array}{cc}\text{L.H.S}& =(\tan \theta +\cot \theta {)}^2\\ & ={(\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta })}^2\\ & ={\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }\right)}^2\end{array}$$

2. Apply the Pythagorean identity.

   The numerator is ${\sin }^2\theta +{\cos }^2\theta =1$.

   $$\begin{array}{cc}\text{L.H.S}& ={\left(\frac{1}{\sin \theta \cos \theta }\right)}^2\\ & =\frac{1}{{\cos }^2\theta }\cdot \frac{1}{{\sin }^2\theta }\end{array}$$

3. Use the reciprocal ratios.

   ${\displaystyle \frac{1}{\cos \theta }}=\sec \theta$ and ${\displaystyle \frac{1}{\sin \theta }}=\text{cosec}\theta$.

   $$\begin{array}{cc}\text{L.H.S}& ={\sec }^2\theta {\text{cosec}}^2\theta \\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Replace the 1 in the numerator.

   From $1+{\tan }^2\theta ={\sec }^2\theta$ we get $1={\sec }^2\theta -{\tan }^2\theta$. Substitute this for the $1$ in the numerator only.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}\\ & =\frac{(\tan \theta +\sec \theta )-({\sec }^2\theta -{\tan }^2\theta )}{\tan \theta -\sec \theta +1}\end{array}$$

2. Factorise the difference of squares.

   Use $a^2-b^2=(a+b)(a-b)$, then take $(\tan \theta +\sec \theta )$ common.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{(\tan \theta +\sec \theta )-(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )}{\tan \theta -\sec \theta +1}\\ & =\frac{(\tan \theta +\sec \theta )[1-\sec \theta +\tan \theta ]}{\tan \theta -\sec \theta +1}\end{array}$$

3. Cancel the common bracket.

   The bracket $1-\sec \theta +\tan \theta$ is the same as the denominator $\tan \theta -\sec \theta +1$.

   $$\begin{array}{cc}\text{L.H.S}& =\frac{(\tan \theta +\sec \theta )(\tan \theta -\sec \theta +1)}{\tan \theta -\sec \theta +1}\\ & =\tan \theta +\sec \theta \\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Factorise the difference of cubes.

   Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=\sin \theta$, $b=\cos \theta$.

   $$\begin{array}{cc}\text{L.H.S}& ={\sin }^3\theta -{\cos }^3\theta \\ & =(\sin \theta -\cos \theta )({\sin }^2\theta +\sin \theta \cos \theta +{\cos }^2\theta )\end{array}$$

2. Apply the Pythagorean identity.

   Group ${\sin }^2\theta +{\cos }^2\theta$ and replace it by $1$.

   $$\begin{array}{cc}\text{L.H.S}& =(\sin \theta -\cos \theta )(1+\sin \theta \cos \theta )\\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.

Solution

1. Write as a difference of cubes.

   ${\sin }^6\theta =({\sin }^2\theta {)}^3$ and ${\cos }^6\theta =({\cos }^2\theta {)}^3$. Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a={\sin }^2\theta$, $b={\cos }^2\theta$.

   $$\begin{array}{cc}\text{L.H.S}& =({\sin }^2\theta {)}^3-({\cos }^2\theta {)}^3\\ & =({\sin }^2\theta -{\cos }^2\theta )[{\sin }^4\theta +{\sin }^2\theta {\cos }^2\theta +{\cos }^4\theta ]\end{array}$$

2. Complete the square inside the bracket.

   Since $a^2+b^2=(a+b{)}^2-2ab$, we have $a^2+ab+b^2=(a+b{)}^2-ab$.

   $$\begin{array}{cc}\text{L.H.S}& =({\sin }^2\theta -{\cos }^2\theta )[({\sin }^2\theta +{\cos }^2\theta {)}^2-{\sin }^2\theta {\cos }^2\theta ]\end{array}$$

3. Apply the Pythagorean identity.

   Replace ${\sin }^2\theta +{\cos }^2\theta$ by $1$.

   $$\begin{array}{cc}\text{L.H.S}& =({\sin }^2\theta -{\cos }^2\theta )(1-{\sin }^2\theta {\cos }^2\theta )\\ & =\text{R.H.S}\end{array}$$

   Hence $\text{L.H.S}=\text{R.H.S}$.