Vardaan Learning Institute
Chapter Solutions: Introduction to Trigonometry
SECTION A: OBJECTIVE TYPE QUESTIONS (1 Mark Each)
1. If $\sin A = 3/4$, then $\cos A$ is:
$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - 9/16} = \sqrt{7/16} = \sqrt{7}/4$.
Correct Option: (b) $\sqrt{7}/4$
2. The value of $\tan 45^\circ + \cot 45^\circ$ is:
$1 + 1 = 2$.
Correct Option: (c) 2
3. If $\sin(A+B) = 1$ and $\cos(A-B) = \sqrt{3}/2$... then A is:
$A+B = 90^\circ$ and $A-B = 30^\circ$. Adding: $2A = 120^\circ \Rightarrow A = 60^\circ$.
Correct Option: (c) $60^\circ$
4. The value of $(\sin 30^\circ + \cos 30^\circ) - (\sin 60^\circ + \cos 60^\circ)$ is:
$\sin 30 = \cos 60 = 1/2$. $\cos 30 = \sin 60 = \sqrt{3}/2$.
$(1/2 + \sqrt{3}/2) - (\sqrt{3}/2 + 1/2) = 0$.
Correct Option: (b) 0
5. The value of $(\sec A + \tan A)(1 - \sin A)$ is:
$(\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A) = \frac{1+\sin A}{\cos A}(1-\sin A) =
\frac{1-\sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A$.
Correct Option: (d) $\cos A$
6. The value of $9 \sec^2 A - 9 \tan^2 A$ is:
$9(\sec^2 A - \tan^2 A) = 9(1) = 9$.
Correct Option: (b) 9
7. If $x = a \cos \theta$ and $y = b \sin \theta$, then the value of $b^2 x^2 + a^2 y^2$
is:
$b^2(a \cos \theta)^2 + a^2(b \sin \theta)^2 = b^2 a^2 \cos^2 \theta + a^2 b^2 \sin^2 \theta$
$= a^2 b^2 (\cos^2 \theta + \sin^2 \theta) = a^2 b^2$.
Correct Option: (b) $a^2 b^2$
8. The value of $(1 + \tan^2 A)(1 - \sin A)(1 + \sin A)$ is:
$(\sec^2 A)(1 - \sin^2 A) = \sec^2 A \cdot \cos^2 A = \frac{1}{\cos^2 A} \cdot \cos^2 A = 1$.
Correct Option: (b) 1
9. If $\sin \alpha = 1/2$ and $\cos \beta = 1/2$, find the value of $\alpha + \beta$.
$\sin \alpha = 1/2 \Rightarrow \alpha = 30^\circ$.
$\cos \beta = 1/2 \Rightarrow \beta = 60^\circ$.
$\alpha + \beta = 30^\circ + 60^\circ = 90^\circ$.
Correct Option: (c) $90^\circ$
10. Assertion (A): The value of $\sin \theta$ increases as $\theta$ increases... Reason
(R): $\sin 0 = 0, \sin 90 = 1$.
As $\theta$ increases from 0 to 90, $\sin \theta$ goes from 0 to 1. R confirms endpoint values. A is
true, R is true and explains the trend.
Correct Option: (a) Both A and R are true and R is the correct explanation of A.
SECTION B: SHORT ANSWER TYPE QUESTIONS (2 Marks Each)
11. Prove that: $(\sec A + \tan A)(1 - \sin A) = \cos A$.
LHS $= (\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A) = \frac{1+\sin A}{\cos A}(1-\sin A)$
$= \frac{1-\sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A =$ RHS.
Proved
12. If $\sin \theta = \cos \theta$, find the value of $2 \tan \theta + \cos^2 \theta$.
$\sin \theta = \cos \theta \Rightarrow \tan \theta = 1 \Rightarrow \theta = 45^\circ$.
Value $= 2(1) + (\frac{1}{\sqrt{2}})^2 = 2 + 0.5 = 2.5$.
2.5
13. Evaluate: $2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$.
$2(1)^2 + (\frac{\sqrt{3}}{2})^2 - (\frac{\sqrt{3}}{2})^2 = 2$.
2
14. Prove that $\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$.
Multiply num and den by $\sqrt{1+\sin A}$.
$= \sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \frac{1+\sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin
A}{\cos A} = \sec A + \tan A$.
Proved
SECTION C: SHORT ANSWER TYPE II QUESTIONS (3 Marks Each)
15. If $\tan A + \sin A = m$ and $\tan A - \sin A = n$, show that $m^2 - n^2 =
4\sqrt{mn}$.
$m^2 - n^2 = (m+n)(m-n) = (2\tan A)(2\sin A) = 4\tan A \sin A$.
$4\sqrt{mn} = 4\sqrt{(\tan A + \sin A)(\tan A - \sin A)} = 4\sqrt{\tan^2 A - \sin^2 A}$
$= 4\sqrt{\frac{\sin^2}{\cos^2} - \sin^2} = 4\sin A \sqrt{\sec^2 - 1} = 4\sin A \tan A$.
LHS = RHS.
Proved
16. If $\sec \theta + \tan \theta = p$, prove that $\sin \theta = \frac{p^2-1}{p^2+1}$.
$\sec^2 - \tan^2 = 1 \Rightarrow (\sec-\tan)(\sec+\tan)=1 \Rightarrow \sec-\tan = 1/p$.
Add: $2\sec = p + 1/p = \frac{p^2+1}{p} \Rightarrow \cos = \frac{2p}{p^2+1}$.
Sub: $2\tan = p - 1/p = \frac{p^2-1}{p} \Rightarrow \sin = \cos \cdot \tan = \frac{2p}{p^2+1} \cdot
\frac{p^2-1}{2p} = \frac{p^2-1}{p^2+1}$.
Proved
17. Prove that $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$.
LHS $= \frac{1+1/\cos A}{1/\cos A} = \frac{(\cos A + 1)/\cos A}{1/\cos A} = 1 + \cos A$.
RHS $= \frac{1-\cos^2 A}{1-\cos A} = \frac{(1-\cos A)(1+\cos A)}{1-\cos A} = 1+\cos A$.
LHS = RHS.
Proved
18. If $7 \sin^2 \theta + 3 \cos^2 \theta = 4$, show that $\tan \theta =
\frac{1}{\sqrt{3}}$.
$7 \sin^2 \theta + 3(1 - \sin^2 \theta) = 4$
$7 \sin^2 \theta + 3 - 3 \sin^2 \theta = 4$
$4 \sin^2 \theta = 1 \Rightarrow \sin^2 \theta = 1/4 \Rightarrow \sin \theta = 1/2$.
$\theta = 30^\circ$.
$\tan 30^\circ = 1/\sqrt{3}$.
Proved
19. Prove that $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.
Expand LHS: $(\sin^2 A + \csc^2 A + 2) + (\cos^2 A + \sec^2 A + 2)$
$= (\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 4$
$= 1 + (1+\cot^2 A) + (1+\tan^2 A) + 4$
$= 1+1+1+4 + \tan^2 A + \cot^2 A = 7 + \tan^2 A + \cot^2 A$.
Proved
20. If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$, show that $\cos \theta - \sin
\theta = \sqrt{2} \sin \theta$.
Squaring given eq: $\cos^2 + \sin^2 + 2\sin\cos = 2\cos^2$.
$1 + 2\sin\cos = 2\cos^2$.
Now consider $(\cos-\sin)^2 = \cos^2+\sin^2 - 2\sin\cos = 1 - (2\cos^2 - 1) = 2 - 2\cos^2 = 2(1-\cos^2)
= 2\sin^2$.
Taking root: $\cos\theta - \sin\theta = \sqrt{2}\sin\theta$.
Proved
SECTION D: LONG ANSWER TYPE QUESTIONS (5 Marks Each)
21. Prove that: $\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$.
Divide Num and Den by $\sin A$:
$\frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A} = \frac{(\cot A + \csc A) - 1}{1 - \csc A + \cot
A}$.
Replace $1 = \csc^2 A - \cot^2 A$ in numerator:
$\frac{(\cot A+\csc A) - (\csc^2 A-\cot^2 A)}{\cot A-\csc A+1} = \frac{(\cot A+\csc A)(1 - (\csc A-\cot
A))}{\cot A-\csc A+1}$
$= \frac{(\cot A+\csc A)(1-\csc A+\cot A)}{1-\csc A+\cot A} = \cot A + \csc A$.
Proved
22. Prove that: $\frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta} =
1 + \sec \theta \csc \theta$.
Convert to sin/cos:
$\frac{\sin/\cos}{(\sin-\cos)/\sin} + \frac{\cos/\sin}{(\cos-\sin)/\cos} =
\frac{\sin^2}{\cos(\sin-\cos)} - \frac{\cos^2}{\sin(\sin-\cos)}$
$= \frac{\sin^3 - \cos^3}{\sin\cos(\sin-\cos)} =
\frac{(\sin-\cos)(\sin^2+\cos^2+\sin\cos)}{\sin\cos(\sin-\cos)}$
$= \frac{1+\sin\cos}{\sin\cos} = \frac{1}{\sin\cos} + 1 = \sec\theta\csc\theta + 1$.
Proved
SECTION E: CASE STUDY (4 Marks)
23. Case Study: Right Triangle (i) Angle A (ii) Length AB (iii) Eval expression
(i) $\tan A = 1 \Rightarrow A = 45^\circ$.
(ii) $\sin 45 = BC/10 \Rightarrow BC = 5\sqrt{2}$. Since $A=45$, $\triangle$ is isosceles,
$AB=BC=5\sqrt{2}$ cm.
(iii) $\sin A \cos C + \cos A \sin C = \sin(A+C)$. In right $\triangle$ at B, $A+C=90$.
$\sin 90^\circ = 1$.
(i) $45^\circ$ (ii) $5\sqrt{2}$ cm (iii) 1