exercise 8.3
1. Express the trigonometric ratios \(\sin A\), \(\sec A\) ,and \(\tan A\) in terms of \(\cot A\).
Solution: We know that,
\(\cot A = \frac{\cos A}{\sin A}\)
\(tan A\) in terms of \(cot A\)
\(tan A = \frac{1}{\cot A}\)
\(\sin A\) in terms of \(\cot A\)
Use the identity:
\(1 + \cot^2 A = \csc^2 A\)
\(\Rightarrow \csc A = \sqrt{1 + \cot^2 A}\)
\(\Rightarrow \sin A = \frac{1}{\csc A} = \frac{1}{\sqrt{1 + \cot^2 A}}\)
\(\sec A\) in terms of \(\cot A\)
Since
\(\cos A = \cot A \cdot \sin A\)
\(\sec A = \frac{1}{\cos A}\)
Then
\(\sec A = \frac{1}{\cot A \cdot \sin A}\)
\(= \frac{1}{\cot A \cdot \frac{1}{\sqrt{1 + \cot^2 A}}}\)
\(= \frac{\sqrt{1 + \cot^2 A}}{\cot A}\)
2. Write all the other trigonometric ratios of \(\angle A\) in terms of \(\sec A\).
Solution: We know that
\(\sec A = \frac{1}{cos A}\Rightarrow \cos A = \frac{1}{\sec A}\)
\(\sin^2 A + \cos^2 A = 1\)
\(\Rightarrow \sin^2 A = 1 – \cos^2 A = 1 – \left( \frac{1}{\sec A} \right)^2\)
\(\Rightarrow\sin^2 A = \frac{\sec^2 A – 1}{\sec^2 A}\)
\(\Rightarrow \sin A = \frac{\sqrt{\sec^2 A – 1}}{\sec A}\)
Now we express all trigonometric ratios in terms of \(\sec A\)
\(\sin A = \frac{\sqrt{\sec^2 A – 1}}{\sec A}\)
\(\cos A = \frac{1}{\sec A}\)
\(\tan A = \sqrt{\sec^2 A – 1}\)
\(\cot A = \frac{1}{\sqrt{\sec^2 A – 1}}\)
\(\csc A = \frac{\sec A}{\sqrt{\sec^2 A – 1}}\)
(3) Choose the correct option. Justify your choice.
(i) \(9\sec^2 A – 9\tan^2 A=\)
Ans: Correct option is (B)=9
Justification: We know that
\(\sec^2 A – \tan^2 A = 1\)
Multiply both sides by 9:
\(9(\sec^2 A – \tan^2 A) = 9 \times 1 = 9\)
Therefore:
\(9\sec^2 A – 9\tan^2 A = 9\)
(ii) \((1 + \tan\theta + \sec\theta)(1 + \cot\theta – \csc\theta)=\)
ans: Correct option is (C)=2
Justification: We know that,
\(\tan\theta = \frac{\sin\theta}{\cos\theta}\)
\(\cot\theta = \frac{\cos\theta}{\sin\theta}\)
\(\sec\theta = \frac{1}{\cos\theta}\)
\(\csc\theta = \frac{1}{\sin\theta}\)
Substitute these values in given equation,
\((1 + \frac{\sin\theta}{\cos\theta} + \frac{1}{\cos\theta})(1 + \frac{\cos\theta}{\sin\theta} – \frac{1}{\sin\theta})\)
\(= \left(\frac{\cos\theta + \sin\theta + 1}{\cos\theta}\right) \cdot \left(\frac{\sin\theta + \cos\theta – 1}{\sin\theta}\right)\)
Multiply numerators:
\(\cos\theta + \sin\theta + 1)(\cos\theta + \sin\theta – 1) \)
\( = cos\theta + \sin\theta)^2 – 1\)
\(= \cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta – 1 \)
\( = 1 + 2\sin\theta\cos\theta – 1 = 2\sin\theta\cos\theta\)
Denominator:
\(\cos\theta \cdot \sin\theta\)
So the expression becomes:
\(\frac{2\sin\theta\cos\theta}{\cos\theta\sin\theta} = 2\)
(iii) \((\sec A + \tan A)(1 – \sin A)=\)
Ans: Correct option is \((D)=\cos A\)
Justification: We have,
\(\sec A = \frac{1}{\cos A}\)
\(\tan A = \frac{\sin A}{\cos A}\)
then
\(\frac{(1 + \sin A)}{\cos A} (1 – \sin A)\)
\(=\frac{1 – \sin^2 A}{\cos A}\)
\(=\frac{\cos^2 A}{\cos A} \)
\( = \cos A\)
(iv) \(\frac{1 + \tan^2 A}{1 + \cot^2 A}\)
Ans: Correct option is \({\tan^2 A}\)
Justification: we have,
\(1 + \tan^2 A = \sec^2 A\)
\(1 + \cot^2 A = \csc^2 A\)
then the expression becomes as given below
\(\frac{\sec^2 A}{\csc^2 A}\)
Now express in terms of sine and cosine:
\(\sec^2 A = \frac{1}{\cos^2 A}\)
\(\csc^2 A = \frac{1}{\sin^2 A}\)
\(\Rightarrow \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A\)
4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) \((\csc \theta – \cot \theta)^2 = \frac{1 – \cos \theta}{1 + \cos \theta}\)
proof:
L.H.S.
\((\csc \theta – \cot \theta)^2\)
\(= \left( \frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta} \right)^2\)
\(= \left( \frac{1 – \cos \theta}{\sin \theta} \right)^2\)
\(= \frac{(1 – \cos \theta)^2}{\sin^2 \theta}\)
R.H.S.
\(\frac{(1 – \cos \theta)}{(1 + \cos \theta})\)
\(= \frac{1 – \cos \theta}{1 + \cos \theta} \cdot \frac{1 – \cos \theta}{1 – \cos \theta}\)
\(= \frac{(1 – \cos \theta)^2}{1 – \cos^2 \theta}\)
\(= \frac{(1 – \cos \theta)^2}{\sin^2 \theta}\)
Therefore, \(L.H.S=R.H.S\)
Hence proved.
(ii) \(\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A\)
Proof: L.H.S
\(= \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}\)
Simplify first term:
\(\frac{\cos A}{1 + \sin A} \cdot \frac{1 – \sin A}{1 – \sin A} = \frac{\cos A (1 – \sin A)}{1 – \sin^2 A}\)
\(= \frac{\cos A (1 – \sin A)}{\cos^2 A} = \frac{1 – \sin A}{\cos A}\)
now we get,
\(\frac{1 – \sin A}{\cos A} + \frac{1 + \sin A}{\cos A} = \frac{2}{\cos A} = 2 \sec A\)
Therefore L.H.S=R.H.S
Hence proved.
(iii) \(\frac{\tan\theta}{1 – \cot\theta} + \frac{\cot\theta}{1 – \tan\theta} = 1 + \sec\theta \csc\theta\). [Hint : Write the expression in terms of \(\sin\theta\) and \(\cos\theta\)]
proof: We know that,
\(\tan\theta = \frac{\sin\theta}{\cos\theta}\)
\(\cot\theta = \frac{\cos\theta}{\sin\theta}\)
\(\cos\theta – \sin\theta = -(\sin\theta – \cos\theta)\)
L.H.S:
\(=\frac{\frac{\sin\theta}{\cos\theta}}{1 – \frac{\cos\theta}{\sin\theta}} + \frac{\frac{\cos\theta}{\sin\theta}}{1 – \frac{\sin\theta}{\cos\theta}}\)
\(=\frac{\frac{\sin\theta}{\cos\theta}}{1- \frac{\cos\theta}{\sin\theta}}+\frac{\frac{\cos\theta}{\sin\theta}}{1- \frac{\sin\theta}{\cos\theta}}\)
\(=\frac{\sin^2\theta}{\cos\theta(\sin\theta – \cos\theta)} + \frac{\cos^2\theta}{\sin\theta-(\sin\theta – \cos\theta)}\)
\(=\frac{\sin^2\theta}{\cos\theta(\sin\theta – \cos\theta)} – \frac{\cos^2\theta}{\sin\theta(\sin\theta – \cos\theta)}\)
\(=\frac{\sin^3\theta – \cos^3\theta}{\cos\theta \sin\theta (\sin\theta – \cos\theta)}\)
Now use the identity:
\((a^3 – b^3)= (a – b)(a^2 + ab + b^2)\)
So:
\(=\frac{(\sin\theta – \cos\theta)(\sin^2\theta + \sin\theta \cos\theta + \cos^2\theta)}{\cos\theta \sin\theta (\sin\theta – \cos\theta)}\)
\(=\frac{\sin^2\theta + \sin\theta \cos\theta + \cos^2\theta}{\cos\theta \sin\theta}\)
Now using this identity: \(\sin^2\theta + \cos^2\theta = 1\)
So,
\(=\frac{1 + \sin\theta \cos\theta}{\cos\theta \sin\theta}\)
Split the fraction:
\(=\frac{1}{\cos\theta \sin\theta} + \frac{\sin\theta \cos\theta}{\cos\theta \sin\theta}\)
\(=\frac{1}{\cos\theta \sin\theta} + 1\)
\(=1 + \frac{1}{\cos\theta \sin\theta}\)
\( = 1 + \sec\theta \csc\theta\)
\(=L.H.S\)
Hence proved.
(iv) \(\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}\). [Hint : Simplify LHS and RHS separately]
proof: Given that,
\(\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}\)
now,
L.H.S: \(\frac{1 + \sec A}{\sec A}\)
\(=\frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}\)
\(=\frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}}\)
\(=\left( \frac{\cos A + 1}{\cos A} \right) \cdot \left( \frac{\cos A}{1} \right)\)
\(=\cos A + 1\)
R.H.S: \(\frac{\sin^2 A}{1 – \cos A}\)
\(=\frac{1 – \cos^2 A}{1 – \cos A}\)
\(=w\frac{(1 – \cos A)(1 + \cos A)}{1 – \cos A}\)
\(= 1 + \cos A\)
Therefore, \(R.H.S = L.H.S\). Hence proved.
(v) \(\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \csc A + \cot A\).
Using this identity \(\csc^2 A = 1 + \cot^2 A\).
proof: Given,
\(\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \csc A + \cot A\)
L.H.S: \(\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1}\)
now dividing numerator and denominator by \(\sin A\)
\(=\frac{\cot A – 1 + \csc A}{\cot A + 1 – \csc A}\)
\(=\frac{\cot A – (\csc^2 A – cot^2 A) + \csc A}{\cot A + 1 – \csc A}\)
\(=\frac{\cot A + \csc A – {(\csc A + \cot A)(\csc A – \cot A)}}{1 + \cot A – \csc A}\)
\(=\frac{(\cot A + \csc A)(1 + \cot A – \csc A)}{(1 + \cot A – \csc A)}\)
\(=\cot A + \csc A \)
= R.H.S
Hence proved.
(vi) \(\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A \)
proof: Given \(\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A \)
squaring both side
\((\sqrt{\frac{1 + \sin A}{1 – \sin A}})^2=(\sec A + \tan A)^2\)
\(\frac{1+\sin A}{1-\sin A}=\sec^2 A +\tan^2 A+2\tan A\sec A\)
now ,
\(R.H.S= \sec^2 A +\tan^2 A+2\tan A\sec A\)
\(=\frac{1}{\cos^2 A}+\frac{\sin^2 A}{\cos^2 A}+2\frac{\sin A}{\cos^2A}\)
\(=\frac{1+\sin^2 A +2\sin A }{\cos^2}\)
\(=\frac{(1+\sin A)^2}{1-\sin^2 A}\)
\(=\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}\)
\(=\frac{1+\sin A}{1-\sin A}\)
\(L.H.S=R.H.S\)
Hence proved.
(vii)\(\frac{\sin \theta – 2\sin^3 \theta}{2\cos^3 \theta – \cos \theta}= \tan \theta\)
proof: Given,
\(\frac{\sin \theta – 2\sin^3 \theta}{2\cos^3 \theta – \cos \theta} = \tan \theta\)
L.H.S
\(=\frac{\sin \theta – 2\sin^3 \theta}{2\cos^3 \theta – \cos \theta}\)
\(=\frac{(\sin \theta)(1- 2\sin^2 \theta)}{(\cos \theta)(2\cos^2 \theta -1)}\)
Now using this identity
\(1 – 2\sin^2 \theta = \cos 2\theta\)
\(2\cos^2 \theta – 1 = \cos 2\theta\)
therefore,
\(=\frac{\sin\theta\cdot\cos2 \theta}{\cos \theta\cdot\cos2 \theta}\)
\(=\frac{\sin \theta}{\sin \theta}\)
\(=\tan \theta\)
\(=R.H.S\)
Hence proved.
(viii)\(\sin A + \csc A)^2 + (\cos A + \sec A)^2=7+\tan^2 A +\cot^2 A\)
proof: Given,
\(\sin A + \csc A)^2 + (\cos A + \sec A)^2=7+\tan^2 A+\cot^2 A\)
Now,
L.H.S.
\(=\sin A + \csc A)^2 + (\cos A + \sec A)^2\)
\(=\sin^2 A+\csc^2 A+2\sin A\csc A+\cos^2 A+\sec^2 A +2\cos A\sec A\)
\(=\sin^2 A+\cos^2 A+2\sin A\csc A+2\cos A\sec A+\csc^2 A+\sec^2 A\)
Using these identities
\(\sin A \csc A = 1\)
\(\cos A \sec A = 1\)
\(\sin^2 A + \cos^2 A = 1\)
\(\csc^2 A = 1 + \cot^2 A\)
\(\sec^2 A = 1 + \tan^2 A\)
Substituting these values we get,
\(=1+2+2+(1+\cot^2 A)+(1+\tan^2 A)\)
\(=7+\cot^2 A +\tan^2 A\)
\(=R.H.S\)
Hence proved.
(ix)\(\csc A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}\).
[Hint : Simplify LHS and RHS separately].
proof: \((\csc A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}\)
L.H.S
\(=(\csc A – \sin A)(\sec A – \cos A)\)
\(=(\frac{1}{\sin A} – \sin A)(\frac{1}{\cos A}-(\cos A)\)
\(=(\frac{1-\sin^2 A}{\sin A})(\frac{1-\cos^2 A}{\cos A})\)
\(= \frac{\cos^2 A \cdot \sin^2 A}{\sin A \cdot \cos A}\)
\(= \sin A \cdot \cos A\)
R.H.S
\(=\frac{1}{\tan A + \cot A}\)
\(=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}\)
\(=\frac{1}{\frac{\sin^2 A+\cos^2A}{\sin A\cos A}}\)
\(=\frac{1}{\frac{1}{\sin A\cos A}}\)
\(=\sin A\cdot\cos A\)
Therefore,\(L.H.S=R.H.S\)
Hence proved.
(x)\((\frac{1 + \tan^2 A}{1 + \cot^2 A})=(\frac{1-\tan A}{1-\cot A})^2= \tan^2 A\)
proof: First we are going to prove that
\(\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A\)
using these identities
\(\tan A=\frac{\sin A}{\cos A}\)
\(\cot A=\frac{\cos A}{\sin A}\)
Now, L.H.S.
\(=\frac{1+\tan^2 A}{1+\cot^2 A}\)
\(=\frac{1+\frac{\sin^2 A}{\cos^2 A}}{1+\frac{\cos^2 A}{\sin^2 A}}\)
\(=\frac{\frac{\cos^2 A+\sin^2 A}{\cos^2 A}}{\frac{\sin^2 A+\cos^2 A}{\sin^2 A}}\)
\(= \frac{\cos^2 A + \sin^2 A}{\cos^2 A} \cdot \frac{\sin^2 A}{\sin^2 A + \cos^2 A}\)
\(= \frac{\sin^2 A}{\cos^2 A}\)
\(= \tan^2 A\)
\(=R.H.S\)
Now we are going to prove that,
\((\frac{1-\tan A}{1-\cot A})^2= \tan^2 A\)
L.H.S
\(=(\frac{1 – \tan A}{1 – \cot A})^2\)
\(=(\frac{1 – \frac{\sin A}{\cos A}}{1 – \frac{\cos A}{\sin A}})^2\)
\(=(\frac{\frac{\cos A – \sin A}{\cos A}}{\frac{\sin A – \cos A}{\sin A}})^2\)
\(=(\frac{\cos A – \sin A}{\cos A} \cdot \frac{\sin A}{\sin A – \cos A})^2\)
\(=(\frac{\cos A – \sin A}{\cos A} \cdot \frac{\sin A}{-(\cos A – \sin A)})^2\)
\(=(-\frac{\sin A}{\cos A})^2\)
\(=(-\tan A)^2\)
\(=\tan^2 A\)
\(=R.H.S\)
Therefore,
\(\frac{1 + \tan^2 A}{1 + \cot^2 A} =(\frac{1 – \tan A}{1 – \cot A})^2=\tan^2 A\)
Hence proved.
