CBSE Class 10 Maths HOTs Introduction to Trigonometry Set 02

VERY SHORT ANSWER TYPE QUESTIONS

Question. Evaluate : \(\tan^2 30^\circ \sin 30^\circ + \cos 60^\circ \sin^2 90^\circ \tan^2 60^\circ - 2 \tan 45^\circ \cos^2 0^\circ \sin 90^\circ\)
Answer: \(= \left(\frac{1}{\sqrt{3}}\right)^2 \times \frac{1}{2} + \frac{1}{2} \times (1)^2 \times (\sqrt{3})^2 - 2 \times 1 \times 1^2 \times 1\)
\(= \frac{1}{3} \times \frac{1}{2} + \frac{1}{2} \times 1 \times 3 - 2 \times 1 \times 1 \times 1\)
\(= \frac{1}{6} + \frac{3}{2} - 2 = \frac{1 + 9 - 12}{6} = \frac{-2}{6} = -\frac{1}{3}\)

Question. Evaluate : \(\sin^2 30^\circ \cos^2 45^\circ + 4 \tan^2 30^\circ + \frac{1}{2} \sin 90^\circ - 2 \cos^2 90^\circ + \frac{1}{24}\)
Answer: \(= \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{\sqrt{2}}\right)^2 + 4 \left(\frac{1}{\sqrt{3}}\right)^2 + \frac{1}{2}(1)^2 - 2(0)^2 + \frac{1}{24}\)
\(= \frac{1}{4}\left(\frac{1}{2}\right) + \frac{4}{3} + \frac{1}{2} + \frac{1}{24} = \frac{1}{8} + \frac{4}{3} + \frac{1}{2} + \frac{1}{24}\)
\(= \frac{3 + 32 + 12 + 1}{24} = \frac{48}{24} = 2\)

Question. Evaluate : \(4(\sin^4 30^\circ + \cos^4 60^\circ) - 3(\cos^2 45^\circ - \sin^2 90^\circ)\)
Answer: \(= 4\left[\left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4\right] - 3\left[\left(\frac{1}{\sqrt{2}}\right)^2 - (1)^2\right]\)
\(= 4\left[\frac{1}{16} + \frac{1}{16}\right] - 3\left[\frac{1}{2} - 1\right]\)
\(= 4 \times \frac{2}{16} - 3 \times \left(-\frac{1}{2}\right)\)
\(= \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2\)

Question. If \(4 \cos \theta = 11 \sin \theta\), find the value of \(\frac{11 \cos \theta - 7 \sin \theta}{11 \cos \theta + 7 \sin \theta}\).
Answer: We have \(4 \cos \theta = 11 \sin \theta\)
or, \(\cos \theta = \frac{11}{4} \sin \theta\)
Now \(\frac{11 \cos \theta - 7 \sin \theta}{11 \cos \theta + 7 \sin \theta} = \frac{11 \times \frac{11}{4} \sin \theta - 7 \sin \theta}{11 \times \frac{11}{4} \sin \theta + 7 \sin \theta}\)
\(= \frac{\sin \theta (\frac{121}{4} - 7)}{\sin \theta (\frac{121}{4} + 7)}\)
\(= \frac{121 - 28}{121 + 28} = \frac{93}{149}\)

Question. Evaluate: \(\frac{3 \tan^2 30^{\circ} + \tan^2 60^{\circ} + \csc 30^{\circ} - \tan 45^{\circ}}{\cot^2 45^{\circ}}\)
Answer: \(\frac{3 \tan^2 30^{\circ} + \tan^2 60^{\circ} + \csc 30^{\circ} - \tan 45^{\circ}}{\cot^2 45^{\circ}}\)
\(= \frac{3 \times (\frac{1}{\sqrt{3}})^2 + (\sqrt{3})^2 + 2 - 1}{(1)^2}\)
\(= \frac{3 \times \frac{1}{3} + 3 + 2 - 1}{1}\)
\(= 1 + 3 + 2 - 1 = 5\)

Question. Express \( \cos 68^\circ + \tan 76^\circ \) in terms of the angles between \( 0^\circ \) and \( 45^\circ \).
Answer: Here we will use \( \cos(90^\circ - \theta) = \sin \theta \) and \( \tan(90^\circ - \theta) = \cot \theta \).
\( \cos 68^\circ + \tan 76^\circ = \cos(90^\circ - 22^\circ) + \tan(90^\circ - 14^\circ) \)
\( = \sin 22^\circ + \cot 14^\circ \)

Question. Find the value of \( \cos 2\theta \), if \( 2 \sin 2\theta = \sqrt{3} \).
Answer: We have \( 2 \sin 2\theta = \sqrt{3} \)
\( \sin 2\theta = \frac{\sqrt{3}}{2} = \sin 60^\circ \)
\( 2\theta = 60^\circ \)
Hence, \( \cos 2\theta = \cos 60^\circ = \frac{1}{2} \).

Question. Find the value of \( \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ \). Is it equal to \( \sin 90^\circ \) or \( \cos 90^\circ \)?
Answer: \( \sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ = \frac{1}{2} \times \frac{1}{2} + \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \)
\( = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1 \)
It is equal to \( \sin 90^\circ = 1 \) but not equal to \( \cos 90^\circ \) as \( \cos 90^\circ = 0 \).

Question. Evaluate : \( \frac{6 \sin 23^\circ + \sec 79^\circ + 3 \tan 48^\circ}{\text{cosec } 11^\circ + 3 \cot 42^\circ + 6 \cos 67^\circ} \)
Answer: \( = \frac{6 \sin(90^\circ - 23^\circ) + \text{cosec}(90^\circ - 79^\circ) + 3 \cot(90^\circ - 48^\circ)}{\text{cosec } 11^\circ + 3 \cot 42^\circ + 6 \cos 67^\circ} \)
\( = \frac{6 \cos 67^\circ + \text{cosec } 11^\circ + 3 \cot 42^\circ}{\text{cosec } 11^\circ + 3 \cot 42^\circ + 6 \cos 67^\circ} \)
\( = 1 \)

Question. If \( \sqrt{3} \sin \theta - \cos \theta = 0 \) and \( 0^\circ < \theta < 90^\circ \), find the value of \( \theta \).
Answer: We have \( \sqrt{3} \sin \theta - \cos \theta = 0 \)
\( \sqrt{3} \sin \theta = \cos \theta \)
\( \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \)
\( \tan \theta = \frac{1}{\sqrt{3}} = \tan 30^\circ \)
\( \theta = 30^\circ \)

Question. Evaluate : \( \frac{\cos 45^\circ}{\sec 30^\circ} + \frac{1}{\sec 60^\circ} \)
Answer: We have \( \frac{\cos 45^\circ}{\sec 30^\circ} + \frac{1}{\sec 60^\circ} = \frac{1/\sqrt{2}}{2/\sqrt{3}} + \frac{1}{2} \)
\( = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{1}{2} \)
\( = \frac{\sqrt{6} + 2}{4} \)

Question. Express \( \cos 71^\circ - \sin 57^\circ + \tan 63^\circ \) in terms of trigonometric ratios of angles between \( 0^\circ \) and \( 45^\circ \).
Answer: \( \cos 71^\circ - \sin 57^\circ + \tan 63^\circ \)
\( = \cos(90^\circ - 19^\circ) - \sin(90^\circ - 33^\circ) + \tan(90^\circ - 27^\circ) \)
\( = \sin 19^\circ - \cos 33^\circ + \cot 27^\circ \)

Question. If \( \cos(40^\circ + x) = \sin 30^\circ \), find the value of \( x \).
Answer: We have \( \cos(40^\circ + x) = \sin 30^\circ \)
\( \cos(40^\circ + x) = \sin(90^\circ - 60^\circ) \)
\( \cos(40^\circ + x) = \cos 60^\circ \)
\( 40^\circ + x = 60^\circ \)
\( x = 20^\circ \)

Question. Evaluate : \(\frac{5 \cos^2 60^\circ + 4 \cos^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 60^\circ}\)
Answer: \(\frac{5 \cos^2 60^\circ + 4 \cos^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 60^\circ}\)
\(= \frac{5\left(\frac{1}{2}\right)^2 + 4\left(\frac{\sqrt{3}}{2}\right)^2 - (1)^2}{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2}\)
\(= \frac{\frac{5}{4} + 3 - 1}{\frac{1}{4} + \frac{1}{4}}\)
\(= \frac{\frac{5}{4} + 2}{\frac{1}{2}} = \frac{\frac{13}{4}}{\frac{1}{2}} = \frac{13}{2}\)

Question. If \(\sin 3\theta = \cos(\theta - 6^\circ)\), where \(3\theta\) and \(\theta - 6^\circ\) are both acute angles, find the value of \(\theta\).
Answer: According to the question,
\(\sin 3\theta = \cos(\theta - 6^\circ)\)
\(\cos(90^\circ - 3\theta) = \cos(\theta - 6^\circ)\)
\(90^\circ - 3\theta = \theta - 6^\circ\)
\(4\theta = 90^\circ + 6^\circ = 96^\circ\)
Thus \(\theta = \frac{96^\circ}{4} = 24^\circ\)

Question. If \(\theta\) be an acute angle and \(5 \operatorname{cosec} \theta = 7\), Then evaluate \(\sin \theta + \cos^2 \theta - 1\).
Answer: We have \(5 \operatorname{cosec} \theta = 7\)
\(\operatorname{cosec} \theta = \frac{7}{5}\)
\(\sin \theta = \frac{5}{7}\) [\(\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}\)]
\(\sin \theta + \cos^2 \theta - 1 = \sin \theta - (1 - \cos^2 \theta)\)
\(= \sin \theta - \sin^2 \theta\) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]
\(= \frac{5}{7} - \left(\frac{5}{7}\right)^2 = \frac{5}{7} - \frac{25}{49} = \frac{35 - 25}{49} = \frac{10}{49}\)

Question. If \(\sin A = \frac{\sqrt{3}}{2}\), find the value of \(2 \cot^2 A - 1\).
Answer: Using \(\because \cot^2 \theta = -1 + \operatorname{cosec}^2 \theta\) we have
\(2 \cot^2 A - 1 = 2(\operatorname{cosec}^2 A - 1) - 1\)
\(= \frac{2}{\sin^2 A} - 3\)
\(= \frac{2}{(\frac{\sqrt{3}}{2})^2} - 3 = \frac{8}{3} - 3 = \frac{-1}{3}\)
Thus \(2 \cot^2 A - 1 = -\frac{1}{3}\)

Question. Prove that : \(\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta\)
Answer: \(\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta - 1)}\)
\(= \frac{\sin \theta(\sin^2 \theta + \cos^2 \theta - 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta - \sin^2 \theta - \cos^2 \theta)}\)
\(= \frac{\tan \theta (\cos^2 \theta - \sin^2 \theta)}{(\cos^2 \theta - \sin^2 \theta)}\)
\(= \tan \theta\) Hence Proved.

Question. Express : \( \sin A, \tan A \) and \( \text{cosec } A \) in terms of \( \sec A \).
Answer: (1) \( \sin^2 A + \cos^2 A = 1 \)
\( \sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{1}{\sec^2 A}} = \sqrt{\frac{\sec^2 A - 1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sec A} \)
(2) \( \tan A = \frac{\sin A}{\cos A} = \sin A \sec A = \frac{\sqrt{\sec^2 A - 1}}{\sec A} \times \sec A = \sqrt{\sec^2 A - 1} \)
(iii) \( \text{cosec } A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}} \)

Question. Prove that : \( \frac{\cot^3 \theta \cdot \sin^3 \theta}{(\cos \theta + \sin \theta)^2} + \frac{\tan^3 \theta \cdot \cos^3 \theta}{(\cos \theta + \sin \theta)^2} = \frac{\sec \theta \text{ cosec } \theta - 1}{\text{cosec } \theta + \sec \theta} \)
Answer: \( \frac{\cot^3 \theta \cdot \sin^3 \theta}{(\cos \theta + \sin \theta)^2} + \frac{\tan^3 \theta \cdot \cos^3 \theta}{(\cos \theta + \sin \theta)^2} \)
\( = \frac{\frac{\cos^3 \theta}{\sin^3 \theta} \times \sin^3 \theta}{(\cos \theta + \sin \theta)^2} + \frac{\frac{\sin^3 \theta}{\cos^3 \theta} \times \cos^3 \theta}{(\cos \theta + \sin \theta)^2} \)
\( = \frac{\cos^3 \theta}{(\cos \theta + \sin \theta)^2} + \frac{\sin^3 \theta}{(\cos \theta + \sin \theta)^2} \)

Question. Prove that : \(\frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} = \frac{\sec \theta + 1}{\sec \theta - 1}\).
Answer: We have \(\frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta} = \frac{\frac{\sin \theta}{\cos \theta} + \sin \theta}{\frac{\sin \theta}{\cos \theta} - \sin \theta}\)
\(= \frac{\sin \theta (\frac{1}{\cos \theta} + 1)}{\sin \theta (\frac{1}{\cos \theta} - 1)}\)
\(= \frac{\sec \theta + 1}{\sec \theta - 1}\)
Hence Proved.

Question. If \(\sin(A + B) = 1\) and \(\sin(A - B) = \frac{1}{2}\), \(0 \le A + B \le 90^{\circ}\) and \(A > B\), then find \(A\) and \(B\).
Answer: We have \(\sin(A + B) = 1 = \sin 90^{\circ}\)
\(A + B = 90^{\circ}\) ...(1)
and \(\sin(A - B) = \frac{1}{2} = \sin 30^{\circ}\)
\(A - B = 30^{\circ}\) ...(2)
Solving eq. (1) and (2), we obtain \(A = 60^{\circ}\) and \(B = 30^{\circ}\)

SHORT ANSWER TYPE QUESTIONS

Question. Find \(\csc 30^{\circ}\) and \(\cos 60^{\circ}\) geometrically.
Answer: Let a triangle \(ABC\) with each side equal to \(2a\).
\(\angle A = \angle B = \angle C = 60^{\circ}\)
Draw \(AD\) perpendicular to \(BC\)
\(\Delta BDA \cong \Delta CDA\) by RHS
\(BD = CD\)
\(\angle BAD = \angle CAD = 30^{\circ}\) by CPCT
\(AD = \sqrt{3}a\)
In \(\Delta BDA\), \(\csc 30^{\circ} = \frac{AB}{BD} = \frac{2a}{a} = 2\)
and \(\cos 60^{\circ} = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2}\)

Question. If \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = \frac{1}{\sqrt{3}} \), \( A+B \leq 90^\circ \), \( A > B \), then find \( A \) and \( B \).
Answer: We have \( \tan(A + B) = \sqrt{3} = \tan 60^\circ \)
\( A + B = 60^\circ \) ...(1)
Again \( \tan(A - B) = \frac{1}{\sqrt{3}} = \tan 30^\circ \)
or, \( A - B = 30^\circ \) ...(2)
Adding equations (1) and (2), we obtain,
\( 2A = 90^\circ \)
\( A = \frac{90^\circ}{2} = 45^\circ \)
Putting this value of \( A \) in equation (1), we get
\( B = 60^\circ - A = 60^\circ - 45^\circ = 15^\circ \)
Hence, \( A = 45^\circ \) and \( B = 15^\circ \)

Question. If \( \cos(A - B) = \frac{\sqrt{3}}{2} \) and \( \sin(A + B) = \frac{\sqrt{3}}{2} \), find \( A \) and \( B \), where \( (A + B) \) and \( (A - B) \) are acute angles.
Answer: We have \( \cos(A - B) = \frac{\sqrt{3}}{2} = \cos 30^\circ \)
\( A - B = 30^\circ \) ...(1)
Also \( \sin(A + B) = \frac{\sqrt{3}}{2} = \sin 60^\circ \)
\( A + B = 60^\circ \) ...(2)
Adding equations (1) and (2), we obtain,
\( 2A = 90^\circ \)
\( A = 45^\circ \)
Putting this value of \( A \) in equation (1), we get \( B = 15^\circ \)

Question. If \( \sin \phi = \frac{1}{2} \), show that \( 3 \cos \phi - 4 \cos^3 \phi = 0 \).
Answer: We have \( \sin \theta = \frac{1}{2} \)
\( \phi = 30^\circ \)
Now substituting this value of \( \phi \) in LHS we have
\( 3 \cos \phi - 4 \cos^3 \phi = 3 \cos 30^\circ - 4 \cos^3 30^\circ \)
\( = 3 \left( \frac{\sqrt{3}}{2} \right) - 4 \left( \frac{\sqrt{3}}{2} \right)^3 \)
\( = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} \)
\( = 0 \)
Hence Proved

Question. If in a triangle \( ABC \) right angled at \( B \), \( AB = 6 \) units and \( BC = 8 \) units, then find the value of \( \sin A \cos C + \cos A \sin C \).
Answer: As per question statement figure is shown below.
We have \( AC^2 = 8^2 + 6^2 = 100 \)
\( AC = 10 \)
Now \( \sin A = \frac{8}{10} \), \( \cos A = \frac{6}{10} \)
and \( \sin C = \frac{6}{10} \), \( \cos C = \frac{8}{10} \)
Thus \( \sin A \cos C + \cos A \sin C = \frac{8}{10} \times \frac{8}{10} + \frac{6}{10} \times \frac{6}{10} \)
\( = \frac{64}{100} + \frac{36}{100} \)
\( = \frac{100}{100} = 1 \)

Question. Simplify : \(\frac{\sin \theta \sec(90^\circ - \theta) \tan \theta}{\text{cosec}(90^\circ - \theta) \cos \theta \cot(90^\circ - \theta)} - \frac{\tan(90^\circ - \theta)}{\cot \theta}\)
Answer: \(\sec(90^\circ - \theta) = \text{cosec } \theta\),
\(\tan(90^\circ - \theta) = \cot \theta\),
\(\cot(90^\circ - \theta) = \tan \theta\),
\(\text{cosec}(90^\circ - \theta) = \sec \theta\)
Hence,
\(\frac{\sin \theta \text{cosec } \theta \tan \theta}{\sec \theta \cos \theta \tan \theta} - \frac{\cot \theta}{\cot \theta}\)
\(= \frac{\sin \theta \times \frac{1}{\sin \theta} \times \tan \theta}{\frac{1}{\cos \theta} \times \cos \theta \times \tan \theta} - 1\)
\(= 1 - 1 = 0\)

Question. Verify : \(\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \frac{\sin \theta}{1 + \cos \theta}\), for \(\theta = 60^\circ\)
Answer: LHS \(= \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \sqrt{\frac{1 - \cos 60^\circ}{1 + \cos 60^\circ}}\)
\(= \sqrt{\frac{1 - \frac{1}{2}}{1 + \frac{1}{2}}} = \sqrt{\frac{\frac{1}{2}}{\frac{3}{2}}} = \frac{1}{\sqrt{3}}\)
RHS \(= \frac{\sin \theta}{1 + \cos \theta} = \frac{\sin 60^\circ}{1 + \cos 60^\circ}\)
\(= \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}\)
RHS = LHS
Hence, relation is verified for \(\theta = 60^\circ\).

Question. If \(\tan A + \cot A = 2\), then find the value of \(\tan^2 A + \cot^2 A\).
Answer: We have \(\tan A + \cot A = 2\)
Squaring both sides, we have
\((\tan A + \cot A)^2 = (2)^2\)
\(\tan^2 A + \cot^2 A + 2 \tan A \cdot \cot A = 4\)
\(\tan^2 A + \cot^2 A + 2 \tan A \times \frac{1}{\tan A} = 4\)
\(\tan^2 A + \cot^2 A + 2 = 4\)
\(\tan^2 A + \cot^2 A = 4 - 2\)
\(\tan^2 A + \cot^2 A = 2\)

Question. If \(\cos \theta + \sin \theta = \sqrt{2} \cos \theta\), show that \(\cos \theta - \sin \theta = \sqrt{2} \sin \theta\).
Answer: We have \(\cos \theta + \sin \theta = \sqrt{2} \cos \theta\)
\(\sin \theta = \cos \theta(\sqrt{2} - 1)\)
\(\sin \theta = \frac{\cos \theta(\sqrt{2} - 1)(\sqrt{2} + 1)}{(\sqrt{2} + 1)}\)
or, \(\sin \theta = \frac{\cos \theta(2 - 1)}{\sqrt{2} + 1}\)
\((\sqrt{2} + 1) \sin \theta = \cos \theta\)
\(\sqrt{2} \sin \theta + \sin \theta = \cos \theta\)
\(\cos \theta - \sin \theta = \sqrt{2} \sin \theta\) Hence proved.

Question. Prove that : \(\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A\).
Answer: LHS \(= \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A}\)
\(= \frac{\cos A}{1 - (\frac{\sin A}{\cos A})} + \frac{\sin A}{1 - (\frac{\cos A}{\sin A})}\)
\(= \frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A}\)
\(= \frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A}\)
\(= \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A}\)
\(= \frac{(\cos A - \sin A)(\cos A + \sin A)}{(\cos A - \sin A)}\)
\(= \cos A + \sin A\)
\(= \sin A + \cos A = \text{RHS}\) Hence proved.

Question. Given that \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\), find the values of \(\tan 75^\circ\) and \(\tan 90^\circ\) by taking suitable values of \(A\) and \(B\).
Answer: (i) \(\tan 75^\circ = \tan(45^\circ + 30^\circ)\)
\(= \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}\)
\(= \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}\)
\(= \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)
\(= \frac{3 + 2\sqrt{3} + 1}{(\sqrt{3})^2 - (1)^2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}\)
Hence \(\tan 75^\circ = 2 + \sqrt{3}\)
(ii) \(\tan 90^\circ = \tan(60^\circ + 30^\circ)\)
\(= \frac{\tan 60^\circ + \tan 30^\circ}{1 - \tan 60^\circ \tan 30^\circ} = \frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{1 - \sqrt{3} \times \frac{1}{\sqrt{3}}} = \frac{\frac{3+1}{\sqrt{3}}}{0} = \infty\)

Question. If \(15 \tan^2 \theta + 4 \sec^2 \theta = 23\), then find the value of \((\sec \theta + \text{cosec } \theta)^2 - \sin^2 \theta\).
Answer: We have \(15 \tan^2 \theta + 4 \sec^2 \theta = 23\)
\(15 \tan^2 \theta + 4(1 + \tan^2 \theta) = 23\) [\(\sec^2 \theta = 1 + \tan^2 \theta\)]
\(15 \tan^2 \theta + 4 + 4 \tan^2 \theta = 23\)
\(19 \tan^2 \theta = 19\)
\(\tan \theta = 1 = \tan 45^\circ \implies \theta = 45^\circ\)
Now, \((\sec \theta + \text{cosec } \theta)^2 - \sin^2 \theta\)
\(= (\sec 45^\circ + \text{cosec } 45^\circ)^2 - \sin^2 45^\circ\)
\(= (\sqrt{2} + \sqrt{2})^2 - \left(\frac{1}{\sqrt{2}}\right)^2 = (2\sqrt{2})^2 - \frac{1}{2} = 8 - \frac{1}{2} = \frac{15}{2}\)

Question. If \( \sqrt{3} \cot^2 \theta - 4 \cot \theta + \sqrt{3} = 0 \), then find the value of \( \cot^2 \theta + \tan^2 \theta \).
Answer: We have \( \sqrt{3} \cot^2 \theta - 4 \cot \theta + \sqrt{3} = 0 \)
Let \( \cot \theta = x \), then we have
\( \sqrt{3}x^2 - 4x + \sqrt{3} = 0 \)
\( \sqrt{3}x^2 - 3x - x + \sqrt{3} = 0 \)
\( (x - \sqrt{3})(\sqrt{3}x - 1) = 0 \)
Thus \( x = \sqrt{3} \) or \( x = \frac{1}{\sqrt{3}} \)
or \( \cot \theta = \sqrt{3} \) or \( \cot \theta = \frac{1}{\sqrt{3}} \)
Therefore \( \theta = 30^\circ \) or \( \theta = 60^\circ \)
If \( \theta = 30^\circ \), then
\( \cot^2 30^\circ + \tan^2 30^\circ = (\sqrt{3})^2 + \left(\frac{1}{\sqrt{3}}\right)^2 \)
\( = 3 + \frac{1}{3} = \frac{10}{3} \)
If \( \theta = 60^\circ \), then
\( \cot^2 60^\circ + \tan^2 60^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 + (\sqrt{3})^2 \)
\( = \frac{1}{3} + 3 = \frac{10}{3} \).

Question. Evaluate : \( \frac{\cos 65^\circ}{\sin 25^\circ} - \frac{\tan 20^\circ}{\cot 70^\circ} - \sin 90^\circ + \tan 5^\circ \tan 35^\circ \tan 60^\circ \tan 55^\circ \tan 85^\circ \).
Answer: We have \( \frac{\cos 65^\circ}{\sin 25^\circ} = \frac{\cos 65^\circ}{\sin(90^\circ - 65^\circ)} = \frac{\cos 65^\circ}{\cos 65^\circ} = 1, \)
\( \frac{\tan 20^\circ}{\cot 70^\circ} = \frac{\tan(90^\circ - 70^\circ)}{\cot 70^\circ} = \frac{\cot 70^\circ}{\cot 70^\circ} = 1 \)
and \( \sin 90^\circ = 1 \)
\( \tan 5^\circ \tan 35^\circ \tan 60^\circ \tan 55^\circ \tan 85^\circ \)
\( = \tan(90^\circ - 85^\circ)\tan(90^\circ - 55^\circ) \cdot \tan 55^\circ \tan 60^\circ \tan 85^\circ \)
\( = \cot 85^\circ \tan 85^\circ \cot 55^\circ \tan 55^\circ \cdot \sqrt{3} \)
\( = 1 \times 1 \times \sqrt{3} = \sqrt{3} \)
Now given expression \( = 1 - 1 - 1 + \sqrt{3} = \sqrt{3} - 1 \)

Question. Prove that : \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \).
Answer: \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \)
\( = \frac{\tan^2 \theta}{\tan \theta - 1} + \frac{1}{(1 - \tan \theta)\tan \theta} \)
\( = \frac{\tan^2 \theta}{\tan \theta - 1} - \frac{1}{(\tan \theta - 1)\tan \theta} \)
\( = \frac{\tan^3 \theta - 1}{(\tan \theta - 1)\tan \theta} \)
[\(\because a^3 - b^3 = (a - b)(a^2 + b^2 + ab)\)]
\( = \frac{(\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1)}{(\tan \theta - 1)\tan \theta} \)
\( = \frac{\tan^2 \theta + \tan \theta + 1}{\tan \theta} \)
\( = \tan \theta + 1 + \cot \theta \)
Hence Proved.

Question. In an acute angled triangle \( ABC \), if \( \sin(A + B - C) = \frac{1}{2} \) and \( \cos(B + C - A) = \frac{1}{\sqrt{2}} \), find \( \angle A, \angle B \) and \( \angle C \).
Answer: We have \( \sin(A + B - C) = \frac{1}{2} = \sin 30^\circ \)
or, \( A + B - C = 30^\circ \) ...(1)
and \( \cos(B + C - A) = \frac{1}{\sqrt{2}} = \cos 45^\circ \)
or, \( B + C - A = 45^\circ \) ...(2)
Adding equation (1) and (2), we get
\( 2B = 75^\circ \)
or, \( B = 37.5^\circ \)
Now subtracting equation (2) from equation (1) we get,
\( 2(A - C) = -15^\circ \)
or, \( A - C = -7.5^\circ \) ...(3)
Now \( A + B + C = 180^\circ \)
\( A + C + 37.5^\circ = 180^\circ \)
\( A + C = 180^\circ - 37.5^\circ = 142.5^\circ \) ...(4)
Adding equation (3) and (4), we have
\( 2A = 135^\circ \)
or, \( A = 67.5^\circ \)
and, \( C = 75^\circ \)
Hence, \( \angle A = 67.5^\circ, \angle B = 37.5^\circ, \angle C = 75^\circ \)

Question. Prove that : \( \sec^4 \theta - \sec^2 \theta = \tan^4 \theta + \tan^2 \theta \)
Answer: To prove \( \sec^4 \theta - \sec^2 \theta = \tan^4 \theta + \tan^2 \theta \)
We have \( \sec^4 \theta - \sec^2 \theta = \sec^2 \theta(\sec^2 \theta - 1) \)
[\( \because 1 + \tan^2 \theta = \sec^2 \theta \)]
\( = \sec^2 \theta(\tan^2 \theta) \)
\( = (1 + \tan^2 \theta)\tan^2 \theta \)
\( = \tan^2 \theta + \tan^4 \theta \)
Hence Proved.

Question. Find the value of \( \theta \), if, \( \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4; \theta \leq 90^\circ \)
Answer: We have \( \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 4 \)
\( \frac{\cos \theta(1 + \sin \theta) + \cos \theta(1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} = 4 \)
\( \frac{\cos \theta[1 + \sin \theta + 1 - \sin \theta]}{1 - \sin^2 \theta} = 4 \)
\( \frac{\cos \theta(2)}{\cos^2 \theta} = 4 \)
\( \frac{1}{\cos \theta} = 2 \)
\( \cos \theta = \frac{1}{2} \)
\( \cos \theta = \cos 60^\circ \)
Thus \( \theta = 60^\circ \)

Question. Prove that : \( -1 + \frac{\sin A \sin(90^\circ - A)}{\cot(90^\circ - A)} = -\sin^2 A \)
Answer: \( -1 + \frac{\sin A \sin(90^\circ - A)}{\cot(90^\circ - A)} = -1 + \frac{\sin A \cos A}{\tan A} \)
\( = -1 + \sin A \cos A \times \cot A \)
\( = -1 + \sin A \cos A \times \frac{\cos A}{\sin A} \)
\( = -1 + \cos^2 A = -(1 - \cos^2 A) \)
\( = -\sin^2 A \) Hence Proved.

Question. Prove that : \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \csc A - \cot A \)
Answer: \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \sqrt{\frac{1 - \cos A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A}} \)
\( = \sqrt{\frac{(1 - \cos A)^2}{(1 - \cos^2 A)}} \)
\( = \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}} \) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]
\( = \frac{1 - \cos A}{\sin A} = \frac{1}{\sin A} - \frac{\cos A}{\sin A} \)
\( = \csc A - \cot A \) Hence Proved.

Question. If \( \sin \theta - \cos \theta = \frac{1}{2} \), then find the value of \( \sin \theta + \cos \theta \).
Answer: We have \( \sin \theta - \cos \theta = \frac{1}{2} \)
Squaring both sides, we get
\( (\sin \theta - \cos \theta)^2 = \left(\frac{1}{2}\right)^2 \)
\( \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta = \frac{1}{4} \)
\( 1 - 2 \sin \theta \cos \theta = \frac{1}{4} \)
\( 2 \sin \theta \cos \theta = 1 - \frac{1}{4} = \frac{3}{4} \)
Again, \( (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta \)
\( = 1 + 2 \sin \theta \cos \theta \)
\( = 1 + \frac{3}{4} = \frac{7}{4} \)
Thus \( \sin \theta + \cos \theta = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \)

Question. Prove that : \(\frac{\cos A}{1 + \tan A} - \frac{\sin A}{1 + \cot A} = \cos A - \sin A\)
Answer: \(\frac{\cos A}{1 + \tan A} - \frac{\sin A}{1 + \cot A}\)
\(= \frac{\cos A}{1 + \frac{\sin A}{\cos A}} - \frac{\sin A}{1 + \frac{\cos A}{\sin A}}\)
\(= \frac{\cos^2 A}{\cos A + \sin A} - \frac{\sin^2 A}{\sin A + \cos A}\)
\(= \frac{\cos^2 A - \sin^2 A}{(\sin A + \cos A)}\)
\(= \frac{(\cos A + \sin A)(\cos A - \sin A)}{\sin A + \cos A}\)
\(= \cos A - \sin A\) Hence Proved.

Question. Prove that : \((\cot \theta - \operatorname{cosec} \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}\)
Answer: To prove \((\cot \theta - \operatorname{cosec} \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}\)
\((\cot \theta - \operatorname{cosec} \theta)^2 = \left(\frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)^2\)
\(= \left(\frac{\cos \theta - 1}{\sin \theta}\right)^2\)
\(= \frac{(1 - \cos \theta)^2}{\sin^2 \theta}\) [Using \((a-b)^2 = (b-a)^2\)]
\(= \frac{(1 - \cos \theta)^2}{(1 - \cos^2 \theta)}\) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]
\(= \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}\)
\(= \frac{1 - \cos \theta}{1 + \cos \theta}\) Hence Proved.

Question. Prove that : \((\operatorname{cosec} \theta - \sin \theta)(\sec \theta - \cos \theta)(\tan \theta + \cot \theta) = 1\)
Answer: [DDE-M, 2015]
LHS \(= (\operatorname{cosec} \theta - \sin \theta)(\sec \theta - \cos \theta)(\tan \theta + \cot \theta)\)
\(= \left(\frac{1}{\sin \theta} - \sin \theta\right)\left(\frac{1}{\cos \theta} - \cos \theta\right)\left(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\right)\)
\(= \left(\frac{1 - \sin^2 \theta}{\sin \theta}\right)\left(\frac{1 - \cos^2 \theta}{\cos \theta}\right)\left(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\right)\)
\(= \frac{\cos^2 \theta}{\sin \theta} \times \frac{\sin^2 \theta}{\cos \theta} \times \left(\frac{1}{\sin \theta \cos \theta}\right)\) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]
\(= \cos \theta \sin \theta \times \frac{1}{\sin \theta \cos \theta} = 1\) Hence Proved.

Question. Show that : \(\operatorname{cosec}^2 \theta - \tan^2 (90^\circ - \theta) = \sin^2 \theta + \sin^2 (90^\circ - \theta)\)
Answer: LHS \(= \operatorname{cosec}^2 \theta - \tan^2 (90^\circ - \theta)\)
\(= \frac{1}{\sin^2 \theta} - \frac{\sin^2(90^\circ - \theta)}{\cos^2(90^\circ - \theta)}\)
\(= \frac{1}{\sin^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta}\)
\(= \frac{1 - \cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} = 1\)
RHS \(= \sin^2 \theta + \sin^2 (90^\circ - \theta)\)
\(= \sin^2 \theta + \cos^2 \theta = 1\)
LHS = RHS. Hence Proved.

Question. Prove that : \(\frac{\operatorname{cosec}^2 \theta}{\operatorname{cosec} \theta - 1} - \frac{\operatorname{cosec}^2 \theta}{\operatorname{cosec} \theta + 1} = 2 \sec^2 \theta\)
Answer: We have
\(\frac{\operatorname{cosec}^2 \theta}{\operatorname{cosec} \theta - 1} - \frac{\operatorname{cosec}^2 \theta}{\operatorname{cosec} \theta + 1} = \operatorname{cosec}^2 \theta \left[ \frac{1}{\operatorname{cosec} \theta - 1} - \frac{1}{\operatorname{cosec} \theta + 1} \right]\)
\(= \operatorname{cosec}^2 \theta \left[ \frac{(\operatorname{cosec} \theta + 1) - (\operatorname{cosec} \theta - 1)}{(\operatorname{cosec} \theta - 1)(\operatorname{cosec} \theta + 1)} \right]\)
\(= \operatorname{cosec}^2 \theta \left[ \frac{2}{\operatorname{cosec}^2 \theta - 1} \right]\)
\(= \frac{1}{\sin^2 \theta} \left[ \frac{2 \sin^2 \theta}{(1 - \sin^2 \theta)} \right]\)
\(= \frac{2}{\cos^2 \theta} = 2 \sec^2 \theta\) Hence Proved

Question. Prove that : \(\frac{1}{\operatorname{cosec} A - \cot A} - \frac{1}{\sin A} = \frac{1}{\sin A} - \frac{1}{\operatorname{cosec} A + \cot A}\)
Answer: To prove: \(\frac{1}{\operatorname{cosec} A - \cot A} + \frac{1}{\operatorname{cosec} A + \cot A} = \frac{1}{\sin A} + \frac{1}{\sin A}\)
LHS \(= \frac{(\operatorname{cosec} A + \cot A) + (\operatorname{cosec} A - \cot A)}{(\operatorname{cosec} A - \cot A)(\operatorname{cosec} A + \cot A)}\)
\(= \frac{2 \operatorname{cosec} A}{\operatorname{cosec}^2 A - \cot^2 A}\)
\(= \frac{2 \operatorname{cosec} A}{1}\) [\(\because \operatorname{cosec}^2 A - \cot^2 A = 1\)]
\(= \frac{2}{\sin A}\)
RHS \(= \frac{1}{\sin A} + \frac{1}{\sin A} = \frac{2}{\sin A}\)
LHS = RHS. Hence Proved.

Question. If \(\sec \theta = x + \frac{1}{4x}\), prove that \(\sec \theta + \tan \theta = 2x\) or \(\frac{1}{2x}\).
Answer: We have \(\sec \theta = x + \frac{1}{4x}\)
Squaring both sides we have
\(\sec^2 \theta = x^2 + \frac{1}{16x^2} + 2 \cdot x \cdot \frac{1}{4x}\)
\(1 + \tan^2 \theta = x^2 + \frac{1}{16x^2} + \frac{1}{2}\)
\(\tan^2 \theta = x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1\)
\(\tan^2 \theta = x^2 + \frac{1}{16x^2} - \frac{1}{2}\)
\(\tan^2 \theta = \left(x - \frac{1}{4x}\right)^2\)
Taking square root both sides we obtain
\(\tan \theta = \pm \left(x - \frac{1}{4x}\right)\)
Case I: Take \(\tan \theta = x - \frac{1}{4x}\)
\(\sec \theta + \tan \theta = \left(x + \frac{1}{4x}\right) + \left(x - \frac{1}{4x}\right) = 2x\)
Case II: Take \(\tan \theta = -\left(x - \frac{1}{4x}\right) = -x + \frac{1}{4x}\)
\(\sec \theta + \tan \theta = \left(x + \frac{1}{4x}\right) + \left(-x + \frac{1}{4x}\right) = \frac{2}{4x} = \frac{1}{2x}\)

HOTS QUESTIONS

Question. Prove that \(\sec^2 \theta + \csc^2 \theta\) can never be less than 2.
Answer: Let \(\sec^2 \theta + \csc^2 \theta = x\)
\(1 + \tan^2 \theta + 1 + \cot^2 \theta = x \implies 2 + \tan^2 \theta + \cot^2 \theta = x\)
Since \(\tan^2 \theta \ge 0\) and \(\cot^2 \theta \ge 0\), thus \(x \ge 2\).
Hence \(\sec^2 \theta + \csc^2 \theta\) can never be less than 2.

Question. (a) Solve for \(\phi\), if \(\tan 5\phi = 1\)
(b) Solve for \(\phi\), if \(\frac{\sin \phi}{1 + \cos \phi} + \frac{1 + \cos \phi}{\sin \phi} = 4\)
Answer: (a) \(\tan 5\phi = 1 \implies \tan 5\phi = \tan 45^{\circ} \implies 5\phi = 45^{\circ} \implies \phi = 9^{\circ}\)
(b) \(\frac{\sin \phi}{1 + \cos \phi} + \frac{1 + \cos \phi}{\sin \phi} = 4 \implies \frac{\sin^2 \phi + (1 + \cos \phi)^2}{\sin \phi(1 + \cos \phi)} = 4\)
\(\frac{\sin^2 \phi + 1 + \cos^2 \phi + 2 \cos \phi}{\sin \phi(1 + \cos \phi)} = 4 \implies \frac{2 + 2 \cos \phi}{\sin \phi(1 + \cos \phi)} = 4\)
\(\frac{2(1 + \cos \phi)}{\sin \phi(1 + \cos \phi)} = 4 \implies \frac{2}{\sin \phi} = 4 \implies \sin \phi = \frac{1}{2} = \sin 30^{\circ}\)
Thus \(\phi = 30^{\circ}\)

Question. If \(\tan A + \sin A = m\) and \(\tan A - \sin A = n\), show that \(m^2 - n^2 = 4\sqrt{mn}\).
Answer: We have \(\tan A + \sin A = m\) and \(\tan A - \sin A = n\)
\(m^2 - n^2 = (\tan A + \sin A)^2 - (\tan A - \sin A)^2\)
= \((\tan^2 A + \sin^2 A + 2\sin A \tan A) - (\tan^2 A + \sin^2 A - 2\sin A \tan A) = 4 \sin A \tan A\)
Now \(4\sqrt{mn} = 4\sqrt{(\tan A + \sin A)(\tan A - \sin A)} = 4\sqrt{\tan^2 A - \sin^2 A}\)
= \(4\sqrt{\frac{\sin^2 A}{\cos^2 A} - \sin^2 A} = 4\sqrt{\frac{\sin^2 A (1 - \cos^2 A)}{\cos^2 A}}\)
= \(4\sqrt{\frac{\sin^2 A \sin^2 A}{\cos^2 A}} = 4 \frac{\sin A \sin A}{\cos A} = 4 \sin A \tan A\)
Thus \(m^2 - n^2 = 4\sqrt{mn}\) Hence Proved