Solutions: Introduction to Trigonometry - Test 1

Q1. (a) \(\frac{3}{4}\)

Given \(\sin A = \frac{3}{5}\). We know \(\sin^2 A + \cos^2 A = 1\).
So, \(\cos^2 A = 1 - (\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}\).
\(\cos A = \sqrt{\frac{16}{25}} = \frac{4}{5}\).
Therefore, \(\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{4}\).

Q2. (b) 0

We know that \(\sin(90^\circ - \theta) = \cos \theta\).
So, \(\sin 42^\circ = \sin(90^\circ - 48^\circ) = \cos 48^\circ\).
Thus, \(\cos 48^\circ - \sin 42^\circ = \cos 48^\circ - \cos 48^\circ = 0\).

Q3. (a) \(\frac{17}{8}\)

Given \(15 \cot A = 8\), so \(\cot A = \frac{8}{15}\).
We can form a right triangle with adjacent = 8, opposite = 15.
Hypotenuse = \(\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\).
So, \(\sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{17}{8}\).

Q4. (c) A is true but R is false.

Assertion (A): For \(0^\circ < A < 90^\circ\), as A increases, \(\sin A\) increases. This is true.
Reason (R): For \(0^\circ < A < 90^\circ\), as A increases, \(\cos A\) increases. This is false.

Q5. If \(\sec \theta + \tan \theta = p\), then prove that \(\sin \theta = \frac{p^2 - 1}{p^2 + 1}\).

Given \(\sec \theta + \tan \theta = p\) --- (1)

We know that \(\sec^2 \theta - \tan^2 \theta = 1\)
\((\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\)
\((\sec \theta - \tan \theta)p = 1\)
\(\sec \theta - \tan \theta = \frac{1}{p}\) --- (2)

Adding (1) and (2):
\(2 \sec \theta = p + \frac{1}{p} = \frac{p^2 + 1}{p}\)
\(\sec \theta = \frac{p^2 + 1}{2p}\)
\(\cos \theta = \frac{2p}{p^2 + 1}\)

Subtracting (2) from (1):
\(2 \tan \theta = p - \frac{1}{p} = \frac{p^2 - 1}{p}\)
\(\tan \theta = \frac{p^2 - 1}{2p}\)

Now, \(\sin \theta = \tan \theta \times \cos \theta = \frac{p^2 - 1}{2p} \times \frac{2p}{p^2 + 1} = \frac{p^2 - 1}{p^2 + 1}\). Hence Proved.

Q6. Prove that \(\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}\).

LHS = \(\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}\)
Divide numerator and denominator by \(\cos \theta\):
\(= \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}\)
\(= \frac{(\tan \theta + \sec \theta) - 1}{(\tan \theta - \sec \theta) + 1}\)
We know \(1 = \sec^2 \theta - \tan^2 \theta\)
\(= \frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{(\tan \theta - \sec \theta) + 1}\)
\(= \frac{(\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{(\tan \theta - \sec \theta) + 1}\)
Take \((\tan \theta + \sec \theta)\) common from numerator:
\(= \frac{(\tan \theta + \sec \theta)[1 - (\sec \theta - \tan \theta)]}{(1 - \sec \theta + \tan \theta)}\)
\(= \frac{(\tan \theta + \sec \theta)(1 - \sec \theta + \tan \theta)}{(1 - \sec \theta + \tan \theta)}\)
\(= \tan \theta + \sec \theta\)
\(= \frac{1}{\sec \theta - \tan \theta}\) (Since \(\sec^2 \theta - \tan^2 \theta = 1 \implies (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\))
\(= \text{RHS}\). Hence Proved.

Q7. If \(\tan(A+B) = \sqrt{3}\) and \(\tan(A-B) = \frac{1}{\sqrt{3}}\); \(0^\circ < A+B \le 90^\circ\); \(A > B\), find A and B.

Given \(\tan(A+B) = \sqrt{3}\)
We know \(\tan 60^\circ = \sqrt{3}\)
So, \(A+B = 60^\circ\) --- (1)

Given \(\tan(A-B) = \frac{1}{\sqrt{3}}\)
We know \(\tan 30^\circ = \frac{1}{\sqrt{3}}\)
So, \(A-B = 30^\circ\) --- (2)

Adding (1) and (2):
\((A+B) + (A-B) = 60^\circ + 30^\circ\)
\(2A = 90^\circ\)
\(A = 45^\circ\)

Substitute A in (1):
\(45^\circ + B = 60^\circ\)
\(B = 60^\circ - 45^\circ\)
\(B = 15^\circ\)
Thus, \(A = 45^\circ\) and \(B = 15^\circ\).

Q8. Prove that: \(\frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A} = 2 \sec A\).

LHS = \(\frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A}\)
\(= \frac{\cos^2 A + (1+\sin A)^2}{(1+\sin A)\cos A}\)
\(= \frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{(1+\sin A)\cos A}\)
\(= \frac{(\cos^2 A + \sin^2 A) + 1 + 2\sin A}{(1+\sin A)\cos A}\)
\(= \frac{1 + 1 + 2\sin A}{(1+\sin A)\cos A}\)
\(= \frac{2 + 2\sin A}{(1+\sin A)\cos A}\)
\(= \frac{2(1 + \sin A)}{(1+\sin A)\cos A}\)
\(= \frac{2}{\cos A}\)
\(= 2 \sec A\)
\(= \text{RHS}\). Hence Proved.

Q9. If \(\sin \theta + \cos \theta = \sqrt{2}\), prove that \(\tan \theta + \cot \theta = 2\).

Given \(\sin \theta + \cos \theta = \sqrt{2}\)
Squaring both sides:
\((\sin \theta + \cos \theta)^2 = (\sqrt{2})^2\)
\(\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 2\)
\(1 + 2\sin \theta \cos \theta = 2\)
\(2\sin \theta \cos \theta = 1\)
\(\sin \theta \cos \theta = 1/2\)

Now consider \(\tan \theta + \cot \theta\):
\(\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}\)
\(= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}\)
\(= \frac{1}{\sin \theta \cos \theta}\)
Substitute \(\sin \theta \cos \theta = 1/2\):
\(= \frac{1}{1/2} = 2\). Hence Proved.

Q10. If \(\tan \theta + \sin \theta = m\) and \(\tan \theta - \sin \theta = n\), prove that \(m^2 - n^2 = 4\sqrt{mn}\).

Given \(m = \tan \theta + \sin \theta\) and \(n = \tan \theta - \sin \theta\).
LHS = \(m^2 - n^2 = (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2\)
Using the identity \((a+b)^2 - (a-b)^2 = 4ab\):
\(m^2 - n^2 = 4 \tan \theta \sin \theta\) --- (1)

RHS = \(4\sqrt{mn}\)
\(= 4\sqrt{(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)}\)
\(= 4\sqrt{\tan^2 \theta - \sin^2 \theta}\)
\(= 4\sqrt{\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta}\)
\(= 4\sqrt{\sin^2 \theta (\frac{1}{\cos^2 \theta} - 1)}\)
\(= 4\sqrt{\sin^2 \theta (\sec^2 \theta - 1)}\)
We know \(\sec^2 \theta - 1 = \tan^2 \theta\):
\(= 4\sqrt{\sin^2 \theta \tan^2 \theta}\)
\(= 4 |\sin \theta \tan \theta|\)
Assuming \(\theta\) is in first quadrant (\(0^\circ < \theta < 90^\circ\)), \(\sin \theta > 0\) and \(\tan \theta > 0\).
\(= 4 \sin \theta \tan \theta\) --- (2)

From (1) and (2), LHS = RHS. Hence Proved.

Q11. Case Study: In a right-angled triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm.

First, find the hypotenuse AC using Pythagoras theorem:
\(AC = \sqrt{AB^2 + BC^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\) cm.

(i) \(\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}\)
\(\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}\). sin A = 7/25, cos A = 24/25 [2]

(ii) \(\sin C = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}\)
\(\cos C = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}\). sin C = 24/25, cos C = 7/25 [1]

(iii) \(\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} = \frac{7}{24}\). tan A = 7/24 [1]