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Example 1. |
Given \(\tan A =\frac {4}{3}\), find the other trigonometric ratios of the angle A. |
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 Let's take a \(\triangle ABC\) Where B is a right angle and \(\angle A\) is a acute angle. We know that \(\tan A = \frac {\text {Opposite side to angle A}}{\text {adjacent side to angle A }}=\frac {BC}{AB}=\frac {4}{3}\) Therefore, if \(BC = 4k\) then \(AB = 3k\). Here k is a positive number. By using Pythagoras Theorem, we can find AC (hypotenuse), \(AC^{2} = AB^{2} + BC^{2}\) \(AC^{2} = (3k)^{2} + (4k)^{2}\) \(AC^{2} = 9k^{2} + 16k^{2}\) \(AC^{2} = 25k^{2}\) \(AC = 5k\) Now that we have all three sides of the right-angled triangle; AC = 5k, AB=3k, BC=4k, so \(\sin A = \frac {BC}{AC}=\frac {4k}{5k}=\frac {4}{5}\) \(\cos A = \frac {AB}{AC}=\frac {3k}{5k}=\frac {3}{5}\) and their reciprocals are \(\text {cosec } A = \frac {1}{\sin A}= \frac {1}{\frac {4}{5}}=\frac {5}{4}\) \(\sec A = \frac {1}{\cos A}= \frac {1}{\frac {3}{5}}=\frac {5}{3}\) \(\cot A = \frac {1}{\tan A}= \frac {1}{\frac {4}{3}}=\frac {3}{4}\) |
Example 2. | If \(\angle B\) and \(\angle Q\) are acute angles such that \(\sin B = \sin Q\), then prove that \(\angle B = \angle Q\). |
 Let's take two triangle ABC and PQR such that \(\sin B = sin Q\) As per construction, we have \(\sin B = \frac {AC}{AB}\) and \(\sin Q = \frac {PR}{PQ}\) As per given, \[\frac {AC}{AB}=\frac {PR}{PQ}\] Therefore, we can say \[\frac {AC}{PR}=\frac {AB}{PQ}=k .........................(i)\] In \(\triangle ABC\) by using Pythagoras theorem: \(AB^{2} = AC^{2} + BC^{2}\) \(BC = \sqrt {AB^{2}-AC^{2}}\) In \(\triangle PQR\) by using Pythagoras theorem: \(PQ^{2} = PR^{2} + QR^{2}\) \(QR = \sqrt {PQ^{2}-PR^{2}}\) So, \(\frac {BC}{QR} = \frac {\sqrt {AB^{2}-AC^{2}}}{\sqrt {PQ^{2}-PR^{2}}} from equation (i) \(\frac {AC}{PR}=k \text { or } AC = k.PR\) \(\frac {AB}{PQ}=k \text { or } AB = k.PQ\) by putting these value So, \(\frac {BC}{QR} = \frac {\sqrt {k^{2}.PQ^{2}-k^{2}.PR^{2}}}{\sqrt {PQ^{2}-PR^{2}}}=\frac {k\sqrt {PQ^{2}-PR^{2}}}{\sqrt {PQ^{2}-PR^{2}}}=k....................... (ii)\) By equation (i) and (ii) we have \(\frac {AC}{PR}=\frac {AB}{PQ}=\frac {BC}{QR}\) We know that two triangles are similar, if their corresponding sides are in the same ratio (or proportion), so we can conclude that triangles ABC and PQR are similar. \(\triangle ABC \sim \triangle PQR\) and therefore \(\angle B = \angle Q\). |
Example 3. | Consider \(\triangle ACB\), right-angled at \(C\), in which \(AB = 29\) units, \(BC = 21\) units and \(angle ABC = \theta\) (see Fig. 8.10). Determine the values of (i) \(\cos^{2}\theta +\sin^{2}\theta\) (ii) \(\cos^{2}\theta -\sin^{2}\theta\) |
 As per shown in figure, In \(\triangle ABC\) , we have \(AC = \sqrt{AB^{2}-BC^{2}}=\sqrt{(29)^{2}-(21)^{2}}\) \(AC = \sqrt{(29-21)(29+21)}=\sqrt{(8)(50)}=\sqrt{400}=\text {20 unit}\) So, By using trigonometric ratio \(\sin \theta = \frac {AC}{AB}=\frac {20}{29}\) \(\cos \theta = \frac {BC}{AB}=\frac {21}{29}\) Hence, (i) \(\cos^{2}\theta +\sin^{2}\theta=\left(\frac {20}{29}\right)^{2}+\left(\frac {21}{29}\right)^{2}=\frac {20^{2}+21^{2}}{29^{2}}=\frac {400+441}{841}=1\) (ii) \(\cos^{2}\theta -\sin^{2}\theta=\left(\frac {21}{29}\right)^{2}-\left(\frac {20}{29}\right)^{2}=\frac {(21+20)(21-20)}{29^{2}}=\frac {41}{841}\) |
Example 4. | In a right triangle \(ABC\), right-angled at \(B\), if tan \(A = 1\), then verify that \(2\sin A \cos A = 1\). |
 As per given, In \(\triangle ABC\), \(\tan = \frac {AB}{BC}=1\) So, we can say that the length of the side opposite \(\angle A\) is equal to the length of the adjacent side. Therefore, \(AB = BC\) Let \(AB = BC = k\), where \(k\) is a positive number. Now, in a right-angled triangle, we can use the Pythagorean theorem: \( AC = \sqrt {AB^{2}+BC^{2}}\) Putting value \( AC = \sqrt {k^{2}+k^{2}}=k\sqrt{2}\) Therefore, \(\sin A = \frac {BC}{AC}=\frac {1}{\sqrt {2}}\) and \(\cos A = \frac {AB}{AC}=\frac {1}{\sqrt {2}}\) Hence, \(2\sin A \cos A = 2\left(\frac {1}{\sqrt {2}}\right)\left(\frac {1}{\sqrt {2}}\right)=1\) |
Example 5. | In \triangle OPQ\), right-angled at \(P\),\(OP = 7\) cm and \(OQ – PQ = 1\) cm (see Fig. 8.12). Determine the values of \(\sin Q\) and \(cos Q\). |
 As per given, In \(\triangle OPQ\), \(OP = 7\) cm \(OQ-PQ = 1\) or \(OQ = 1+PQ\) Now, in a right-angled triangle, we can use the Pythagorean theorem: \(OQ^{2}=OP^{2}+PQ^{2}\) As per given, putting \(OQ\) value \((1+PQ)^{2}=OP^{2}+PQ^{2}\) We know that \((a+b)^{2}=a^{2}+b^{2}+2ab\) so, \(1^{2}+PQ^{2}+2PQ=OP^{2}+PQ^{2}\) Divide both side by \(PQ^{2}\) and \(OP\) value as per given, we get \(1+2PQ = 7^{2}\) \(2PQ = 49-1\) \(2PQ = 48=24\) cm. So, \(PQ = 24 \text { cm}\) and \(OQ = 1+PQ = 25 \text { cm}\) Therefore, \(\sin Q = \frac {OP}{OQ}=\frac {7}{25}\) \(\cos Q = \frac {PQ}{OQ}=\frac {24}{25}\) |
Example 6. | In \(\triangle ABC\), right-angled at \(B\), \(AB = 5\text{ cm}\) and \(\angle ACB = 30^{0}\) (see Fig. 8.19). Determine the lengths of the sides \(BC\) and \(AC\). |
 \(\text{In } \triangle ABC\), right-angled at \(B\), \(AB = 5\, \text{cm}\) and \(\angle ACB = 30^\circ\). We have to find the lengths of the sides \(BC\) and \(AC\). 1. By using Sine Ratio: \( \sin 30^\circ = \frac{AB}{AC} \) by putting \(\sin 30^\circ = \frac {1}{2}\) \( \frac{1}{2} = \frac{5}{AC} \) \( AC = \frac{5}{1/2} = 10\, \text{cm} \) 2. By using Cosine Ratio: \( \cos 30^\circ = \frac{BC}{AC} \) by putting \(\cos 30^\circ = \frac {\sqrt {3}}{2}\) \( \frac{\sqrt{3}}{2} = \frac{BC}{10} \) \( BC = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3}\, \text{cm} \) Therefore, the lengths of the sides \(BC\) and \(AC\) are \(5\sqrt{3}\, \text{cm}\) and \(10\, \text{cm}\), respectively. |
Example 7. | In \(\triangle PQR\), right-angled at \(Q\) (see Fig. 8.20), \(PQ = 3 \text { cm}\) and \(PR = 6\text { cm}\). Determine \(\angle QPR\) and \(\angle PRQ \). |
 Given: \(\text{In } \triangle PQR\), right-angled at \(Q\), \(PQ = 3 \, \text{cm}\) and \(PR = 6 \, \text{cm}\). We have to find \(\angle QPR\) and \(\angle PRQ\). 1. By using Sine Ratio: We know that \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \) putting value \( \sin \angle PRQ = \frac{PQ}{PR} = \frac{3}{6} = \frac{1}{2} \) To find \(\angle PRQ\), take the inverse sine (arcsin) of \(\frac{1}{2}\): \( \angle PRQ = \arcsin\left(\frac{1}{2}\right) \) In degrees, this is equal to \(30^\circ\). 2. We know that sum of angle in a right angle triangle is \(180^\circ\). so \(\angle PQR + \angle QPR + \angle PRQ = 180^\circ \) Putting value \(90^\circ + \angle QPR + 30^\circ = 180^\circ \) \(\angle QPR = 60^\circ \) Therefore, in \(\triangle PQR\), right-angled at \(Q\): \( \angle QPR = 60^\circ \) \( \angle PRQ = 30^\circ \) |
Example 8. | If \(\sin (A-B)=\frac{1}{2}\), \(\cos (A+B)=\frac {1}{2}\), \(0^\circ < A+B \leq 90^\circ\), \(A > B\), Find \(A\) and \(B\). |
Given: \(\sin (A-B)=\frac{1}{2}\), \(\cos (A+B)=\frac {1}{2}\) and \(0^\circ < A+B \leq 90^\circ\) We know that- \(\sin 30^\circ = \frac {1}{2}\) So, we can write as \(\sin (A-B)=\sin 30^\circ \) \(A-B= 30^\circ ...................... (i)\) We also know that- \(\cos 60^\circ = \frac {1}{2}\) So, we can write as \(\cos (A+B)=\cos 60^\circ \) \(A+B= 60^\circ ...................... (i)\) By adding equation \((i)\) and \((ii)\) \(A-B+A+B = 30 + 60\) \(2A = 90\) \(A = 45^\circ\) By putting \(A=45^\circ\) in equation (i) \(45^\circ-B= 30^\circ \) \(45^\circ-30^\circ= B \) \( B = 15^\circ \) So the solution is \(A=45^\circ\) and \(B=15^\circ\) which satisfying given condition. |
Example 9. | Express the ratios \(\cos A\), \(\tan A\) and \(\sec A\) in terms of \(\sin A\). |
We know that, Pythagorean identity is: \(\cos^{2}A + \sin^{2}A = 1\) or \(\cos^{2}A = 1-\sin^{2}A\) or \(\cos A = \pm \sqrt {1-\sin^{2}A}\) Here A is acute angle when \(A<90|) so, it is a positive angle, then \(\cos A = \sqrt {1-\sin^{2}A}...............(i)\) Ratio of \(\bf \sec A\) \(\sec A = \frac {1}{\cos A}\) Putting \(\cos A\) value from equation \((i)\) \(\sec A = \frac {1}{\sqrt {1-\sin^{2}A}}\) Ratio of \(\bf \tan A\) \(\tan A = \frac {\sin A}{\cos A}\) Putting \(\cos A\) value from equation \((i)\) \(\tan A = \frac {\sin A}{{\sqrt {1-\sin^{2}A}}}\) |
Ex. 10. | Prove that \(\sec A (1-\sin A)(\sec A + tan A) = 1\). |
We know that \(\sec A = \frac {1}{\cos A}\) and \(\tan A =\frac {sin A}{cos A}\) \(\sec A (1-\sin A)(\sec A + \tan A)\) Putting \(\sec A\) and \(\tan A value\) \(= \frac{1}{\cos A} (1 - \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) \) \( =\frac{(1 - \sin A)}{\cos A} \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) \) \( =\frac{(1 - \sin A)}{\cos A} \left(\frac{1+ sin A}{\cos A} \right) \) \( =\frac{(1 - \sin A)(1+ sin A)}{\cos A\times \cos A} \) \( =\frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A} \) Since, we know that \(a^{2}-b^{2}=(a-b)(a+b)\) \( =\frac{1 - \sin^2 A}{\cos^2 A} \) We also know that \(\cos^{2} A+\sin^{2} A = 1\) or \(\cos^{2} A= 1-\sin^{2} A\) So we can write as \( =\frac{\cos^2 A}{\cos^2 A} \) \( =1\) Therefore, \(\sec A (1-\sin A)(\sec A + \tan A) = 1\). Its Proved. |
Ex. 11. | Prove that \(\frac {\cot A - \cos A}{\cot A + \cos A} = \frac {\text {cosec }A-1}{\text {cosec }A+1}\). |
\begin{align*}
\text{LHS} &= \frac{\cot A - \cos A}{\cot A + \cos A} \\
&= \frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A} \\
&= \frac{\frac{\cos A - \sin A \cdot \cos A}{\sin A}}{\frac{\cos A + \sin A \cdot \cos A}{\sin A}} \\
&= \frac{\cos A(1 - \sin A)}{\cos A(1 + \sin A)} \\
&= \frac{1 - \sin A}{1 + \sin A}\\ &= \frac {\frac {(1-\sin A)}{\sin A}}{\frac {(1+\sin A)}{\sin A}} \\ =& \frac {\frac {1}{\sin A}-\frac {\sin A}{\sin A}}{\frac {1}{\sin A}+\frac {\sin A}{Sin A}} \\ &= \frac {\frac {1}{sin A}-1}{\frac {1}{sin A}+1} \\ &= \frac {\text {cosec }A-1} {\text {cosec }A+1}.
\end{align*}
Therefore, \(\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text {cosec }A - 1}{\text {cosec }A + 1}\). |
Ex. 12. | Prove that \(\frac {\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac {1}{\sec\theta-\tan\theta}\), using the identity \(\sec^{0}\theta = 1+tan^{0}\theta\). |
\begin{align*}
\text{LHS} &=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} \\
&= \left(\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}\right)\times \left(\frac{\sin\theta+\cos\theta+1}{\sin\theta+\cos\theta+1}\right) \\
&= \left(\frac{\sin\theta+1-\cos\theta}{\sin\theta+\cos\theta-1}\right)\times \left(\frac{\sin\theta+1+\cos\theta}{\sin\theta+\cos\theta+1}\right) \\
&= \frac {(\sin\theta+1)^{2}-\cos^{2}\theta}{(\sin\theta+\cos\theta)^{2}-1^{2}}\\
&=\frac {\sin^{2}\theta+1+2\sin\theta-\cos^{2}\theta}{\sin^{2}\theta+\cos^{2}\theta+2\sin\theta\cos\theta-1}\\
&= \text {We know that } \sin^{2}\theta+\cos^2\theta = 1 \text { and } \sin^{2}\theta=1-\cos^2\theta \\
&= \text {Therefore,}\\
&= \frac {1-\cos^{2}\theta+1+2\sin\theta-\cos^{2}\theta}{1+2\sin\theta\cos\theta-1}\\
&= \frac {2-2\cos^{2}\theta+2\sin\theta}{2\sin\theta\cos\theta}\\
&= \frac {\sin^{2}\theta+\sin\theta}{\sin\theta\cos\theta}\\
&= \frac {\sin\theta+1}{\cos\theta}\\
&= \frac {1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\\
&= \sec\theta + \tan\theta \\
&= \left(\sec\theta + \tan\theta\right) \times \left(\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}\right)\\
&=\frac {\sec^{2}\theta-\tan^{2}\theta}{\sec\theta-\tan\theta}\\
&=\text {As we know and given that} \sec^{2}\theta-\tan^{2}\theta=1 \\
&= \text {Therefore,}\\
&=\frac {1}{\sec\theta-\tan\theta}=RHS\\
&= \text {Hence, It's proved.}
\end{align*}
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