CBSE Class 10 Chapter 8 Introduction of Trigonometry Exercise 8.1

Q1.

In \(\triangle ABC\), right-angled at \(B\), \(AB=24 \text { cm}\), \(BC=7\text{ cm}\). Determine: \(\text{(i)} \sin A, \cos A\) \(\text{(ii)} \sin C, \cos C\)

Ans.

Given: In \(\triangle ABC\)  \(\angle B = 90^{0}\) \(AB = 24 \text { cm}\) \(BC = 7 \text { cm}\) \(AC\) be the hypotenuse By using Pythagoras Theorem: \( AC^2 = AB^2 + BC^2 \) \( AC^2 = 24^2 + 7^2 \) \( AC^2 = 576 + 49 \) \( AC^2 = 625 \, \text{cm}^2 \) \( AC = \sqrt{625} = 25 \) Therefore, \(AC = 25 \, \text{cm}\) (i) \(\sin(A)\), \(\cos(A)\): \( \sin(A) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25} \) \( \cos(A) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25} \) (ii) \(\sin(C)\), \(\cos(C)\): \( \sin(C) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25} \) \( \cos(C) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25} \)

Q2.

In Fig. 8.13, find \(\tan P – \cot R\).

Ans.

Given: In given \(\triangle PQR\)  \(\angle Q = 90^{0}\) \(PQ = 12 \text { cm}\) \(PR = 13 \text { cm}\) By using Pythagoras Theorem: \( PR^2 = QR^2 + PQ^2 \) \( 13^2 = QR^2 + 12^2 \) \( 169 = QR^2 + 144 \) \( 169-144 = QR^2  \) \( QR^2 = 25  \) \( QR = \sqrt{25} = 5 \) Therefore, \(QR = 5 \, \text{cm}\) \(\tan P = \frac {\text {opposite side to}\angle P}{\text {Adjacent side to}\angle P}=\frac{QR}{PQ}\frac {5}{12} .......... (i)\) \(\tan R = \frac {\text {opposite side to}\angle R}{\text {Adjacent side to}\angle R}=\frac{PQ}{QR}\frac {12}{5} .......... (ii)\) We know that \(\cot R\) is reciprocal of \(\tan R\), so \(\cot R = \frac {1}{\tan R}\) Putting value of \(\tan R\) from equation (ii) \(\cot R = \frac {1}{\frac {12}{5}}=\frac {5}{12}....... (iii)\) Thus,  \(\tan P – \cot R\) Putting value from equation (i) and (iii) \(=\frac{5}{12}-\frac {5}{12}=0\)

Q3.

\(\text {If} \sin A = \frac {3}{4} \text {, calculate }\cos A \text { and } \tan A\).

Ans.

Given: In given \(\triangle ABC\)  \(\angle B = 90^{0}\) \(\sin A = \frac {\text {Opposite side to } \angle A}{Hypotenuse}=\frac {BC}{AC}=\frac {3}{4}\) Let \(BC = 3k \) \(AC = 4k \) By using Pythagoras Theorem: \( AC^2 = AB^2 + BC^2 \) \( 4k^2 = AB^2 + 3k^2 \) \( 16k^{2} = AB^2 + 9k^{2} \) \( 16k^{2}-9k^{2} = AB^2  \) \(  AB^2 =7k^2\) \( AB = \sqrt{7k^2} = \sqrt{7}k \) Therefore, \(AB =\sqrt{7}\) \(\cos A = \frac {\text {Adjacent side of } \angle A}{Hypotenuse}=\frac {AB}{AC}=\frac {\sqrt{7}k}{4k}=\frac {\sqrt{7}}{4}\) \(\tan A = \frac {\text {Opposite side of } \angle A}{\text {Adjacent side of} \angle A}=\frac {BC}{AB}=\frac {3k}{\sqrt {7}k}=\frac {3}{\sqrt {7}}\) Alternative Method: If \(\sin A = \frac {3}{4}\), we can use the Pythagorean identity to find \(\cos A\) and then use the definition of \(\tan A\). The Pythagorean identity is given by: \(\sin^{2} A + \cos^{2} A = 1\) \(\cos^{2} A = 1 - \sin^{2} A\) By putting \(\sin A\) value: \(\cos^{2} A = 1 - \left(\frac {3}{4}\right)^{2}\) \(\cos^{2} A = 1 - \left(\frac {9}{16}\right)\) \(\cos^{2} A = \frac {7}{16}\) \(\cos A = \pm \frac {\sqrt{7}}{4}\) Since, \(\sin A\) is positive and \(\cos A\) is in the same quadrant as \(\sin A\), we take the positive root. \(\cos A = \frac {\sqrt{7}}{4}\) As we know that \(\tan A = \frac {\sin A}{\cos A}\) by putting value \(\tan A = \frac {\frac {3}{4}}{\frac {\sqrt {4}}{4}}\) \(\tan A = \frac {3}{4}\times \frac {4}{\sqrt {7}}\) \(\tan A = \frac {3}{\sqrt {7}}\)

Q4.

Given \(15 \cot A = 8\), find \(\sin A\) and \(sec A\).

Ans.

Given: \(15 \cot A = 8\) \(\cot A = \frac {8}{15}\) We know that  \(\tan A = \frac {1}{\cot A}\) By putting \(\cos A\) value \(\tan A = \frac {1}{\frac {8}{15}}=\frac {15}{8}\) So, \(\tan A = \frac {\text {Opposite side of }\angle A}{\text {Adjacent side of }\angle A}=\frac {15}{8}\) For find AC (Hypotenuse) we use Pythagoras Theorem: \( AC^2 = AB^2 + BC^2 \) \( AC^2 = 8k^2 + 15k^2 \) \( AC^{2} = 64k^2 + 225k^{2} \) \( AC^{2}=289k^{2} \) \( AC = 17k \) Therefore, \(AC =17k\) \(\sin A = \frac {\text {Opposite side to }\angle A}{Hypotenuse}=\frac {BC}{AC}=\frac {15k}{17k}=\frac {15}{17}\) \(\cos A = \frac {\text {Adjacent side to }\angle A}{Hypotenuse}=\frac {AB}{AC}=\frac {8k}{17k}=\frac {8}{17}\) We know that- \(\sec A = \frac {1}{\cos A}= \frac {1}{\frac {8}{17}}=\frac {17}{8}\)

Q5.

Given \(\sec\theta = \frac {13}{12}\), calculate all other trigonometric ratios.

Ans.

Given: \(\sec\theta = \frac {13}{12}\) We know that  \(\cos\theta = \frac {1}{\sec\theta}\) By putting \(\sec \theta\) value \(\cos\theta = \frac {1}{\frac {13}{12}}=\frac {12}{13}\) We also know that Pythagorean identity is: \(\sin^{2} \theta + \cos^{2} \theta = 1\) By putting \(\cos\theta\) value \(\sin^{2} \theta + \left(\frac {12}{13}\right)^{2} = 1\) \(\sin^{2} \theta =1- \frac {144}{169}\) \(\sin^{2} \theta = \frac {25}{169}\) \(\sin \theta = \pm \frac {5}{13}\) Since, \(\sec\theta >0\) in the given expression and \(\sin\theta\) is positive. so, \(\sin \theta = \frac {5}{13}\) The other trigonometric ratios are: \(\tan\theta =\frac {\sin\theta}{\cos\theta}=\frac {\frac {5}{13}}{\frac {12}{13}}=\frac {5}{12}\) \(\cot\theta = \frac {1}{\tan\theta}=\frac {1}{\frac{5}{12}}=\frac {12}{5}\) \(\text {cosec}\theta = \frac {1}{\sin\theta}=\frac {1}{\frac{5}{13}}=\frac {13}{5}\) \(\sec\theta = \frac {13}{12}\text { given}\)

Q6.

If \(\angle A\) and \(\angle B\) are acute angles such that \(\cos A = cos B\), then show that \(\angle A = \angle B\).

Ans.

Given : \(\angle A\) and \(\angle B\) are acute angles. \(\cos A = cos B\) We have to show that \(\angle A = \angle B\). In right angle \(\triangle ABC\) \(\cos A = \frac {Adjacent side to \angle A}{Hypotenuse}=\frac {AC}{AB}\) \(\cos B = \frac {Adjacent side to \angle B}{Hypotenuse}=\frac {BC}{AB}\) As per given that \(\cos A = \cos B\) Therefore,  \(\frac {AC}{AB}=\frac {BC}{AB}\) \(\text {Cross multiply}\) \(AC \times AB = BC \times AB\) \(\text {divide both side by AB, we get}\) \(AC = BC\) Because both ratios are equal, then the lengths of AC and BC are the equal and \(\triangle ABC\) is a Isosceles triangle, then angles opposite those equal sides would be equal. So, \(\angle A = \angle B\) Alternative solution: Let's take a triangle ABC in which CP ┴ AB As per given \(\cos A = \cos B\) \(\frac {AP}{AC}=\frac {BP}{BC}\) \(\frac {AP}{BP}=\frac {AC}{BC}\) Let \(\frac {AP}{BP}=\frac {AC}{BC}=x\) Then, \(AP = x.BP ............ (i)\) \(AC = x.BC ............ (ii)\) By using Pythagoras theorem, we get From \(\triangle CAP\) \(AC^{2} = AP^{2}+CP^{2}\) \(CP^{2}=AC^{2}-AP^{2}........ (iii)\) From \(\triangle CBP\) \(BC^{2}=BP^{2}+CP^{2}\) \(CP^{2}=BC^{2}-BP^{2}......... (iv)\) from equatio (iii) and (iv), we get \(AC^{2}-AP^{2}=BC^{2}-BP^{2}\) Putting value from (i) and (ii) \((x.BC)^{2}-(x.BP)^{2}=BC^{2}-BP^{2}\) \(x^{2}BC^{2}-x^{2}BP^{2}=BC^{2}-BP^{2}\) \(x^{2}(BC^{2}-BP^{2})=BC^{2}-BP^{2}\) \(x^{2}=\frac {BC^{2}-BP^{2}}{BC^{2}-BP^{2}}=1\) \(x=1\) Putt this value in equation (ii) \(AC = 1.BC=BC \) Because angles opposite those equal sides would be equal. So, \(\angle A = \angle B\)

Q7.

\(\text {If } \cot\theta =\frac {7}{8}, \text {evalute;}\) \(\text {(i)} \frac {(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}\) \(\text {(ii)} \cot^{2}\theta\)

Ans.

\(\text {Let's take a } \triangle ABC \text{, where } \angle B \text {is a right angle as per above figure.}\) \(\text {We know that-}\) \(\cot\theta = \frac {\text {Adjacent side to} \theta}{\text {Opposite side to}\theta}=\frac {AB}{BC}=\frac {7}{8}\) \(\text {Let}\)\( AB = 7x\) \(\text {and}\) \(BC = 8x\), \(\text {Here }\)\( x\) \(\text {is a positive integer.}\) By Pythagoras theorem in \(\triangle ABC\) \(AC^{2}=AB^{2}+BC^{2}\) \(AC^{2}=(7x)^{2}+(8x)^{2}\) \(AC^{2}=49x^{2}+64x^{2}\) \(AC^{2}=113x^{2}\) \(AC=\sqrt {113x^{2}}\) \(AC=\sqrt {113x}\) Therefore, \(\sin\theta =\frac {Opposite side to \theta}{Hypotenuse}=\frac {BC}{AC}=\frac {8x}{\sqrt{113x}}=\frac {8}{\sqrt{113}}\) \(\cos\theta =\frac {Adjacent side to \theta}{Hypotenuse}=\frac {AB}{AC}=\frac {7x}{\sqrt{113x}}=\frac {7}{\sqrt{113}}\) \(\text {(i)} \frac {(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}\) \(\text {We know that-} (a+b)(a-b)=a^{2}-b^{2}\) so, \(\frac {(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\frac{1-\sin^{2}\theta}{1-\cos^{2}\theta}\) \(=\frac {1-\left(\frac{8}{\sqrt{113}}\right)^{2}}{1-\left(\frac {7}{\sqrt{113}}\right)^{2}}\) \(=\frac {1-\frac{64}{113}}{1-\frac {49}{113}}\) \(=\frac {\frac {49}{113}}{\frac {64}{113}}\) \(=\frac {49}{64}\) \(\text {(ii)} \cot^{2}\theta\) \(\cot^{2}\theta = \left(\frac {7}{8}\right)^{2}=\frac {49}{64}\)

Q8.

If \(3\cot A = 4\), check whether \(\frac {1-\tan^{2}A}{1+\tan^{2}A}=\cos^{2}A-\sin^{2}A\) or not.

Ans.

Given that: \(3\cot A = 4\)  or \(\cot A = \frac {4}{3}\) We know that- \(\tan A = \frac {1}{\cot A}=\frac {1}{\frac {4}{3}}=\frac {3}{4}\) \(\text {Let's take a } \triangle ABC \text{, where } \angle B \text {is a right angle as per above figure.}\) \(\text {We know that-}\) \(\cot A = \frac {\text {Adjacent side to }\angle A}{\text {Opposite side to}\angle A}=\frac {AB}{BC}=\frac {4}{3}\) \(\text {Let}\)\( AB = 4x\) \(\text {and}\) \(BC = 3x\), \(\text {Here }\)\( x\) \(\text {is a positive integer.}\) By Pythagoras theorem in \(\triangle ABC\) \(AC^{2}=AB^{2}+BC^{2}\) \(AC^{2}=(4x)^{2}+(3x)^{2}\) \(AC^{2}=16x^{2}+9x^{2}\) \(AC^{2}=25x^{2}\) \(AC=\sqrt {25x^{2}}\) \(AC=\sqrt {5x}\) Therefore, \(\sin A =\frac {\text {Opposite side to} \angle A}{Hypotenuse}=\frac {BC}{AC}=\frac {3x}{5x}=\frac {3}{5}\) \(\cos A =\frac {\text {Adjacent side to} \angle A}{Hypotenuse}=\frac {AB}{AC}=\frac {4x}{5x}=\frac {4}{5}\) So, \(L.H.S. = \frac {1-\tan^{2}A}{1+\tan^{2}A}\) \(=\frac{1-\left(\frac {3}{4}\right)^{2}}{1+\left(\frac {3}{4}\right)^{2}}=\frac {1-\frac {9}{16}}{1+\frac {9}{16}}=\frac {16-9}{16+9}=\frac {7}{25}\) \(R.H.S. = \cos^{2}A-\sin^{2}A\) \(\left(\frac {4}{5}\right)^{2}-\left(\frac {3}{5}\right)^{2}=\frac {16}{25}-\frac {9}{25}=\frac {16-9}{25}=\frac {7}{25}\) Hence, \(\frac {1-\tan^{2}A}{1+\tan^{2}A}=\cos^{2}A-\sin^{2}A\)

Q9.

In triangle ABC, right-angled at B, if \(\tan A =\frac {1}{\sqrt {3}}\), find the value of: \(\text {(i)} \sin A \cos C + \cos A\sin C\) \(\text {(ii)} \cos A \cos C + \sin A\sin C\)

Ans.

Given that: \(triangle ABC\) is a right-angled at B. \(\tan A =\frac {1}{\sqrt {3}}\) We know that- \(\tan A = \frac {\text {Opposite side of }\angle A}{\text {Adjacent side of }\angle A}=\frac {BC}{AB}=\frac {1}{\sqrt {3}}\) \(\text {Let}\)\( BC = x\) \(\text {and}\) \(AB = \sqrt{3}x\), \(\text {Here }\)\( x\) \(\text {is a positive integer.}\) By Pythagoras theorem in \(\triangle ABC\) \(AC^{2}=AB^{2}+BC^{2}\) \(AC^{2}=(\sqrt {3}x)^{2}+ (x)^{2}\) \(AC^{2}=3x^{2}+ x^{2}\) \(AC^{2}=4x^{2}\) \(AC=\sqrt {4x^{2}}\) \(AC=2x\) Therefore, \(\sin A = \frac {\text {Opposite side to}\angle A}{\text {Hypotenuse}}=\frac {BC}{AC}=\frac {1}{2}\) \(\cos A = \frac {\text {Adjacent side to}\angle A}{\text {Hypotenuse}}=\frac {AB}{AC}=\frac {\sqrt{3}}{2}\) \(\sin C = \frac {\text {Opposite side to}\angle C}{\text {Hypotenuse}}=\frac {AB}{AC}=\frac {\sqrt{3}}{2}\) \(\cos C = \frac {\text {Adjacent side to}\angle C}{\text {Hypotenuse}}=\frac {BC}{AC}=\frac {1}{2}\) \(\text {(i)} \sin A \cos C + \cos A\sin C\) By putting the value of the trigonometric function: \(=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4}+\frac {3}{4}=\frac {1+3}{4}=\frac {4}{4}=1\) \(\text {(ii)} \cos A \cos C + \sin A\sin C\) By putting the value of the trigonometric function: \(=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac {\sqrt {3}}{4}-\frac {\sqrt {3}}{4}=0\)

Q10.

In \(\triangle PQR\), right-angled at \(Q\), \(PR + QR = 25\)\(\text { cm}\) and \(PQ = 5\) \(\text { cm}\). Determine the values of \(\sin P\), \(\cos P\) and \(tan P\).

Ans.

Given: In \(\triangle PQR\), \(\angle Q\) is right-angle. \(PQ = 5 \text { cm}\) \(PR + QR = 25 \text { cm}\) We have to find \(PR\), \(QR\) and \(\sin P\), \(\cos P\) and \(\tan P\). Let, In \(\triangle PQR\) \(PR = x \text { cm}\) Therefore, \(PR + QR = 25 \text { cm}\) \(x + QR = 25 \text { cm}\) \(QR = (25-x) \text { cm}\) By using Pythagoras theorem: \(PR^{2}=PQ^{2}+QR^{2}\) Putting value \(x^{2}=(5)^{2}+(25-x)^{2}\) by using \((a-b)^{2}=a^{2}-2ab+b^{2}\) \(x^{2}=25+625-50x+x^{2}\) \(x^{2}-x^{2}+50x=650\) \(50x = 650\) \(x=\frac {650}{50}=13 \text { cm}\) So, \(PR = 13 \text {cm}\) \(QR = 25-13 = 12 \text {cm}\) Hence, \(\sin P = \frac {\text {Opposite side to}\angle P}{Hypotenuse}=\frac {QR}{PR}=\frac {12}{13}\) \(\cos P = \frac {\text {Adjacent side to}\angle P}{Hypotenuse}=\frac {PQ}{PR}=\frac {5}{13}\) \(\tan P = \frac {\text {Opposite side to}\angle P}{\text {Adjacent side to}\angle P}=\frac {QR}{PQ}=\frac {12}{5}\)

Q11.

State whether the following are true or false. Justify your answer. (i) The value of \(tan A\) is always less than \(1\). (ii) \(\sec A=\frac {12}{5}\) for some value of angle \(A\). (iii) \(\cos A\) is the abbreviation used for the cosecant of angle \(A\). (iv) \(\cot A\) is the product of \(\cot\) and \(A\). (v) \(\sin\theta = \frac {4}{3}\),for some angle \(\theta\).

Ans.

(i) False, because sides of a right-angled triangle may have any length and also the tangent function \((tan)\) can take values greater than 1. So tan A may have any value. For example : \(tan (60^{0})=\sqrt {3}\) which is greater then 1. (ii) True, Because \(\sec A\) is the secant function, which is the reciprocal of the cosine function. For example : If \(\cos A = \frac {5}{12}\) then \(\sec A = \frac {1}{\cos A}=\frac {1}{\frac {5}{12}}=\frac {12}{5}\). Hence \(\sec A > 1\). (iii) False, because \(\cos A\) represents the cosine function, not the cosecant. The cosecant is represented by \(\text {cosec A}\). (iv) False, because \(cot A\) is the cotangent function evaluated at \(\angle A\). It is not the product of \(cot\) and \(A\). (v) False, Because the sine function can only take values between \(-1\) amd \(1\), so \(\sin\theta = \frac {4}{3}\) is not possible for any \(\angle\theta\).