CBSE Class 10 Mathematics Trigonometry Assignment Set 09

Question. Find the value of \( \sin 75^\circ \) from the relation \( \sin(A+B) = \sin A \cos B + \cos A \sin B \)
Answer: To find the value of \( \sin 75^\circ \), we can set \( A = 45^\circ \) and \( B = 30^\circ \).
Using the given relation:
\( \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \)
\( \sin 75^\circ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \)
\( \sin 75^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Question. Find the value of \( \cos 75^\circ \) from the relation \( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
Answer: To find the value of \( \cos 75^\circ \), we can set \( A = 45^\circ \) and \( B = 30^\circ \).
Using the given relation:
\( \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ \)
\( \cos 75^\circ = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \)
\( \cos 75^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

Question. Find the value of \( \tan 15^\circ \) from the relation \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
Answer: To find the value of \( \tan 15^\circ \), we can set \( A = 45^\circ \) and \( B = 30^\circ \).
Using the given relation:
\( \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \)
\( \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \)
Rationalizing the denominator:
\( \tan 15^\circ = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \)

Question. Prove the following:
(i) \( \sin 90^\circ = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \)
(ii) \( \cos 90^\circ = \cos 60^\circ \cos 30^\circ - \sin 60^\circ \sin 30^\circ \)
Answer:
(i) LHS = \( \sin 90^\circ = 1 \)
RHS = \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{3}{4} + \frac{1}{4} = 1 \)
Since LHS = RHS, \( \sin 90^\circ = \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \) is proved.

(ii) LHS = \( \cos 90^\circ = 0 \)
RHS = \( \cos 60^\circ \cos 30^\circ - \sin 60^\circ \sin 30^\circ = \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \)
Since LHS = RHS, \( \cos 90^\circ = \cos 60^\circ \cos 30^\circ - \sin 60^\circ \sin 30^\circ \) is proved.

Question. If \( A = 45^\circ \), verify \( \tan 90^\circ = \frac{2\tan 45^\circ}{1-\tan^2 45^\circ} \)
Answer:
LHS = \( \tan 90^\circ \), which is undefined.
RHS = \( \frac{2\tan 45^\circ}{1-\tan^2 45^\circ} = \frac{2(1)}{1-1^2} = \frac{2}{0} \), which is undefined.
Since both LHS and RHS are undefined, the relation is verified for \( A = 45^\circ \).

Question. Find the value of \( \theta \) (\( 0^\circ \le \theta \le 90^\circ \)) from the following:
(i) \( \sin 3\theta = 1 \)
(ii) \( 2\cos 3\theta = 1 \)
(iii) \( (\sec \theta - 2)(\tan^3 \theta - 1) = 0 \)
Answer:
(i) \( \sin 3\theta = 1 \implies 3\theta = 90^\circ \implies \theta = 30^\circ \)
(ii) \( 2\cos 3\theta = 1 \implies \cos 3\theta = \frac{1}{2} \implies 3\theta = 60^\circ \implies \theta = 20^\circ \)
(iii) \( (\sec \theta - 2)(\tan^3 \theta - 1) = 0 \)
Either \( \sec \theta - 2 = 0 \implies \sec \theta = 2 \implies \cos \theta = \frac{1}{2} \implies \theta = 60^\circ \)
Or \( \tan^3 \theta - 1 = 0 \implies \tan^3 \theta = 1 \implies \tan \theta = 1 \implies \theta = 45^\circ \)
Thus, the possible values of \( \theta \) are \( 45^\circ, 60^\circ \).

Question. Find \( A \) and \( B \) such that:
(i) \( \tan(A+B) = \sqrt{3} \), \( \tan(A-B) = \frac{1}{\sqrt{3}} \)
(ii) \( \cos(2A-B) = \frac{1}{2} \), \( \sin(A+2B) = \frac{\sqrt{3}}{2} \)
Answer:
(i) From \( \tan(A+B) = \sqrt{3} \), we have:
\( A+B = 60^\circ \) — (Equation 1)
From \( \tan(A-B) = \frac{1}{\sqrt{3}} \), we have:
\( A-B = 30^\circ \) — (Equation 2)
Adding Equation 1 and Equation 2:
\( 2A = 90^\circ \implies A = 45^\circ \)
Substituting \( A = 45^\circ \) in Equation 1:
\( 45^\circ + B = 60^\circ \implies B = 15^\circ \)
Thus, \( A = 45^\circ \) and \( B = 15^\circ \).

(ii) From \( \cos(2A-B) = \frac{1}{2} \), we have:
\( 2A-B = 60^\circ \) — (Equation 1)
From \( \sin(A+2B) = \frac{\sqrt{3}}{2} \), we have:
\( A+2B = 60^\circ \) — (Equation 2)
Multiplying Equation 1 by 2:
\( 4A-2B = 120^\circ \) — (Equation 3)
Adding Equation 2 and Equation 3:
\( 5A = 180^\circ \implies A = 36^\circ \)
Substituting \( A = 36^\circ \) in Equation 2:
\( 36^\circ + 2B = 60^\circ \implies 2B = 24^\circ \implies B = 12^\circ \)
Thus, \( A = 36^\circ \) and \( B = 12^\circ \).

Question. If \( \sin\theta + \cos\theta = \sin\theta - \cos\theta \), Find \( \theta \).
Answer:
Subtracting \( \sin\theta \) from both sides of the equation:
\( \cos\theta = -\cos\theta \)
\( 2\cos\theta = 0 \)
\( \cos\theta = 0 \)
For \( 0^\circ \le \theta \le 90^\circ \), \( \theta = 90^\circ \).

Question. If \( \sin\theta = \cos\theta \), Find \( \theta \) and Prove \( \sin^2\theta + \cos^2\theta = 1 \)
Answer:
Given \( \sin\theta = \cos\theta \). Dividing both sides by \( \cos\theta \) (where \( \cos\theta \ne 0 \)):
\( \tan\theta = 1 \implies \theta = 45^\circ \)

To prove \( \sin^2\theta + \cos^2\theta = 1 \), we substitute \( \theta = 45^\circ \):
LHS = \( \sin^2 45^\circ + \cos^2 45^\circ \)
LHS = \( \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1 = \text{RHS} \).
Hence proved.

Question. Show by means of examples that:
(i) \( \sin(A+B) \ne \sin A + \sin B \)
(ii) \( \cos(A-B) \ne \cos A - \cos B \)
Answer:
(i) Let \( A = 30^\circ \) and \( B = 60^\circ \).
LHS = \( \sin(30^\circ + 60^\circ) = \sin 90^\circ = 1 \)
RHS = \( \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1+\sqrt{3}}{2} \approx 1.366 \)
Since \( 1 \ne 1.366 \), LHS \( \ne \) RHS, proving that \( \sin(A+B) \ne \sin A + \sin B \).

(ii) Let \( A = 60^\circ \) and \( B = 30^\circ \).
LHS = \( \cos(60^\circ - 30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \)
RHS = \( \cos 60^\circ - \cos 30^\circ = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1-\sqrt{3}}{2} \approx -0.366 \)
Since \( 0.866 \ne -0.366 \), LHS \( \ne \) RHS, proving that \( \cos(A-B) \ne \cos A - \cos B \).