INTRODUCTION TO TRIGONOMETRY-Exercise 8.2

1. Evaluate the following :

Solution:

i. $$\begin{aligned}\sin 60^{0}\times \cos 30^{0}+\sin 30^{0}\times &\cos 60^{\circ }\\ \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}&=1\\ \dfrac{3}{4}+\dfrac{1}{4}&=1\\ &=1\end{aligned}$$

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ii. $$\begin{aligned}2\times \tan ^{2}45^{\circ }+\cos ^{2}30^{0}-&\sin ^{2}60^{\circ }\\ 2\cdot 1+\left( \dfrac{\sqrt{3}}{2}\right) ^{2}-\left( \dfrac{\sqrt{3}}{2}\right) ^{2}&=2\\ &=2\end{aligned}$$

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iii. $$\begin{aligned}\dfrac{\cos 45^{0}}{\sec 30^{\circ }+co\sec 30^{0}}\\\\ \cos 45^{\circ }&=\dfrac{1}{\sqrt{2}}\\\\ \sec 30^{\circ }&=\dfrac{1}{\cos 30^{\circ }}\\ &=\dfrac{1}{\dfrac{\sqrt{3}}{2}}\\ &=\dfrac{2}{\sqrt{3}}\\\\ \text{cosec }30^{\circ }&=\dfrac{1}{\sin 30^{\circ }}\\ &=\dfrac{1}{\dfrac{1}{2}}\\ &=\dfrac{2}{1}\\ &=2\end{aligned}$$

Now First calculate \((\sec 30^{\circ }+\text{cosec }30^{\circ })\)

$$\begin{aligned}\sec 30^{\circ }&+\text{cosec }30^{\circ }\\ \dfrac{2}{\sqrt{3}}+2\\ &=\dfrac{2+2\sqrt{3}}{\sqrt{3}}\end{aligned}$$

And then calculate \(\left(\dfrac{\cos 45^{\circ }}{\sec 30^{\circ }+\text{cosec }30^{0}}\right)\)

$$\scriptsize\begin{aligned}&\dfrac{\cos 45^{\circ }}{\sec 30^{\circ }+\text{cosec }30^{0}}\\\\ &=\dfrac{1}{\dfrac{\sqrt{2} \left( 2+2 \sqrt{3}\right) }{\sqrt{3}}}\\\\ &=\dfrac{\sqrt{3}}{2 \sqrt{2}+2 \sqrt{6}}\\\\ &=\dfrac{\sqrt{3} \left( 2 \sqrt{2}-2 \sqrt{6}\right) }{\left( 2 \sqrt{2}+2 \sqrt{6}\right) \left( 2 \sqrt{2}-2 \sqrt{6}\right) }\\\\ &=\dfrac{2 \sqrt{3} \left( \sqrt{2}-\sqrt{6}\right) }{4\times 2-4\times 6}\\\\ &=\dfrac{2 \sqrt{3} \left( \sqrt{2}-\sqrt{6}\right) }{8-24}\\\\ &=\dfrac{2 \sqrt{3} \left( \sqrt{2}-\sqrt{6}\right) }{16}\\\\ &=\dfrac{\sqrt{3} \left( \sqrt{2}-\sqrt{6}\right) }{-8}\\\\ &=\dfrac{+\sqrt{18}-\sqrt{6}}{8}\\\\ &=\dfrac{3 \sqrt{2}-\sqrt{6}}{8}\end{aligned}$$

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iv. $$\begin{aligned}\dfrac{\sin 30^{0}+\tan 45^{0}-\cos ec60^{\circ }}{\sec 30^{0}+\cos 60^{0}+\cot 45^{0}}\\\\ \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}\end{aligned}$$

Calculate the Numerator First

$$\begin{aligned}&\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}} \\\\ &=\dfrac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}\\\\ &=\dfrac{3 \sqrt{3}-4}{2 \sqrt{3}}\\\\ &=\dfrac{3 \sqrt{3}-4}{2 \sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}\\\\ &=\dfrac{9-4 \sqrt{3}}{6}\end{aligned}$$

Then calculate Denominator

$$\begin{aligned}&\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1 \\\\ &=\dfrac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}\\\\ &=\dfrac{4+3 \sqrt{3}}{2 \sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}\\\\ &=\dfrac{4 \sqrt{3}+3\times 3}{2\times 3}\\\\ &=\dfrac{9+4 \sqrt{3}}{6}\end{aligned}$$

Write Numerator & Denominator together and simplify (Rationalize)

$$\begin{aligned}&\dfrac{\dfrac{9-4 \sqrt{3}}{6}}{\dfrac{9+4 \sqrt{3}}{6}}\\\\ &=\dfrac{9-4 \sqrt{3}}{9+4 \sqrt{3}}\times \dfrac{9-4 \sqrt{3}}{9-4 \sqrt{3}}\\\\ &=\dfrac{\left( 9-4 \sqrt{3}\right) ^{2}}{9^{2}-\left( 4 \sqrt{3}\right) ^{2}}\\\\ &=\dfrac{81+16\times 3-72 \sqrt{3}}{81-48}\\\\ &=\dfrac{81+48-72 \sqrt{3}}{33}\\\\ &=\dfrac{129-72 \sqrt{3}}{33}\\\\ &=\dfrac{43-24 \sqrt{3}}{11}\end{aligned}$$

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v. $$\scriptsize\begin{aligned}&\dfrac{5\times \cos ^{2}60^{0}+4\sec ^{2}30^{0}-\tan ^{2}45^{0}}{\sin ^{2}30^{0}+\cos ^{2}30^{\circ }}\\\\ &=\dfrac{5\cdot \left( \dfrac{1}{2}\right) ^{2}+4\cdot \left(\dfrac{2}{\sqrt{3}}\right)^{2}+1}{\left( \dfrac{1}{2}\right) ^{2}+\left( \dfrac{\sqrt{3}}{2}\right) ^{2}}\\\\ &=\dfrac{5\cdot \left( \dfrac{1}{4}\right) +4\cdot \left( \dfrac{4}{3}\right) -1}{\dfrac{1}{4}+\dfrac{3}{4}}\\\\ &=\dfrac{\dfrac{5}{4}+\dfrac{16}{3}-1}{1}\\\\ &=\dfrac{15+64-12}{12}\\\\ &=\dfrac{67}{12}\end{aligned}$$

Q2. Choose the correct option and justify your choice :

Solution:

i. $$\begin{aligned} &\dfrac{2\times \tan 30^{\circ }}{1+\tan ^{2}30^{0}}\\\\ &=\dfrac{2\cdot \dfrac{1}{\sqrt{3}}}{1+\left( \dfrac{1}{\sqrt{3}}\right) ^{2}}\\\\ &=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}\\\\ &=\dfrac{2}{\sqrt{3}\times \left( 1+\dfrac{1}{3}\right) }\\\\ &=\dfrac{2}{\sqrt{3}\times \left( \dfrac{3+1}{3}\right) }\\\\ &=\dfrac{2\times 3}{\sqrt{3}\times \left( 4\right) }\\\\ &=\dfrac{\sqrt{3}}{2}\\\\ \dfrac{\sqrt{3}}{2}&=\sin 60^{\circ }\end{aligned}$$

(A) sin 60° is correct

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ii. $$\begin{aligned}&\dfrac{1-\tan ^{2}45^{0}}{1+\tan ^{2}45^{\circ }}\\ &=\dfrac{1-1}{1+1}\\ &=\dfrac{0}{2}\\ &=0\end{aligned}$$

(D) 0 is correct Answer

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iii. $$\begin{aligned}\sin 2A=2\times \sin A\\ A=0\\ \sin 0=\sin 0\end{aligned}$$

(A) = 0° is correct Answer

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iv. $$\begin{aligned} &\dfrac{2\times \tan 30^{\circ }}{1-\tan ^{2}30^{\circ }}\\ &=\dfrac{2\cdot \dfrac{1}{\sqrt{3}}}{1-\left( \dfrac{1}{\sqrt{3}}\right) ^{2}}\\ &=\dfrac{2}{\sqrt{3}\times \left( 1-\dfrac{1}{3}\right) }\\ &=\dfrac{2}{\sqrt{3}\times \left( \dfrac{3-1}{3}\right) }\\ &=\dfrac{2\times 3}{\sqrt{3}\times 2}\\ &=\sqrt{3}\\ \sqrt{3}&=\tan 60^{\circ }\end{aligned}$$

(C) tan 60° is correct Answer

Q.3

Solution:

$$\begin{align}\tan \left( A+B\right) =\sqrt{3}\\ \Rightarrow A+B=60^{\circ }\tag{1}\\ \lim \left( A-B\right) =\dfrac{1}{\sqrt{3}}\\ \Rightarrow A-B=30^{\circ }\tag{2}\end{align}$$

Adding equation 1- and equation 2

$$\begin{aligned}A+B+\left( A-B\right) &=60^{0}+30^{\circ }\\ A+B+A-B&=90^{\circ }\\ 2A&=90^{\circ }\\ A&=\dfrac{90^{\circ }}{2}\\ A&=45^{\circ }\\ A+B&=60^{\circ }\\ 45^{\circ }+B&=60^{\circ }\\ B&=60^{\circ }-45^{0}\\ B&=15^{\circ }\end{aligned}$$

A = 45°, B = 15°

Q4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of \(\sin \theta\) increases as \(\theta\) increases.
(iii) The value of \(\cos \theta\) increases as \(\theta\) increases. (iv) \(\sin \theta = \cos \theta\) for all values of . (v) cot A is not defined for A = 0°.

Solution:

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