Class 10 Maths: Introduction to Trigonometry – Extra Questions with Answers

Ans: D

Solution:
$\sin{\theta} \cos{(90^{\circ}-\theta)} + \sin{(90^{\circ}-\theta)} \cos{\theta} = \sin{\theta} \sin{\theta} + \cos{\theta} \cos{\theta} = \sin^2{\theta} + \cos^2{\theta} = 1$
Answer: 1

Ans: D

Solution:
In $\triangle ABC$, we have $A + B + C = 180^\circ$.
Since $B = 60^\circ$, we get $A + C = 180^\circ – 60^\circ = 120^\circ$.
Then $\frac{A+C}{2} = \frac{120^\circ}{2} = 60^\circ$.
Therefore, $\tan\left(\frac{A+C}{2}\right) = \tan(60^\circ) = \sqrt{3}$.
Answer: $\sqrt{3}$

Ans: D

Solution:
(i) $\tan 47^\circ \tan 23^\circ \tan 43^\circ \tan 67^\circ = \tan (45^\circ + 2^\circ) \tan (45^\circ – 22^\circ) \tan (45^\circ – 2^\circ) \tan (45^\circ + 22^\circ)$
$= \tan 47^\circ \tan 43^\circ \tan 23^\circ \tan 67^\circ = \tan(45^\circ+2^\circ)\tan(45^\circ-2^\circ)\tan(45^\circ-22^\circ)\tan(45^\circ+22^\circ)$
$= \tan 47^\circ \tan (90^\circ – 47^\circ) \tan 23^\circ \tan (90^\circ – 23^\circ)$
$= \tan 47^\circ \cot 47^\circ \tan 23^\circ \cot 23^\circ$
$= (\tan 47^\circ \cdot \frac{1}{\tan 47^\circ}) (\tan 23^\circ \cdot \frac{1}{\tan 23^\circ}) = 1 \cdot 1 = 1$

(ii) $\cos 37^\circ \cos 53^\circ – \sin 37^\circ \sin 53^\circ = \cos 37^\circ \cos (90^\circ – 37^\circ) – \sin 37^\circ \sin (90^\circ – 37^\circ)$
$= \cos 37^\circ \sin 37^\circ – \sin 37^\circ \cos 37^\circ = 0$

Answer:
(i) 1
(ii) 0

Ans: C

Solution: Square both sides of the equation $\sin\theta + \cos\theta = \frac{7}{5}$ to get $(\sin\theta + \cos\theta)^2 = (\frac{7}{5})^2$. Expanding the left side gives $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = \frac{49}{25}$. Since $\sin^2\theta + \cos^2\theta = 1$, we have $1 + 2\sin\theta\cos\theta = \frac{49}{25}$. Therefore, $2\sin\theta\cos\theta = \frac{49}{25} – 1 = \frac{24}{25}$, and $\sin\theta\cos\theta = \frac{24}{50} = \frac{12}{25}$.

Answer: $\frac{12}{25}$

Ans: A

Solution:
Given $4 \cot B = 3$, we have $\cot B = \frac{3}{4}$. Then $\tan B = \frac{1}{\cot B} = \frac{4}{3}$.
Also, $\frac{1 – \tan^2 B}{1 + \tan^2 B} = \frac{1 – (\frac{4}{3})^2}{1 + (\frac{4}{3})^2} = \frac{1 – \frac{16}{9}}{1 + \frac{16}{9}} = \frac{-\frac{7}{9}}{\frac{25}{9}} = -\frac{7}{25}$.
Since $\cos^2 B – \sin^2 B = \cos(2B)$ and $\frac{1 – \tan^2 B}{1 + \tan^2 B} = \cos(2B)$, we have $\cos^2 B – \sin^2 B = -\frac{7}{25}$.
Therefore, $\frac{1 – \tan^2 B}{1 + \tan^2 B} – (\cos^2 B – \sin^2 B) = -\frac{7}{25} – (-\frac{7}{25}) = 0$.

Answer: 0

Ans: D

Solution:
Square both sides of $2\sin\alpha + 3\cos\alpha = 1$:
$(2\sin\alpha + 3\cos\alpha)^2 = 1^2$
$4\sin^2\alpha + 12\sin\alpha\cos\alpha + 9\cos^2\alpha = 1$ (1)
Let $x = 3\sin\alpha – 2\cos\alpha$.
Square $x$:
$x^2 = (3\sin\alpha – 2\cos\alpha)^2$
$x^2 = 9\sin^2\alpha – 12\sin\alpha\cos\alpha + 4\cos^2\alpha$ (2)
Multiply (1) by $\frac{1}{1}$ and (2) by $\frac{1}{1}$:
$(4\sin^2\alpha + 12\sin\alpha\cos\alpha + 9\cos^2\alpha) + (9\sin^2\alpha – 12\sin\alpha\cos\alpha + 4\cos^2\alpha) = 1 + x^2$
$13\sin^2\alpha + 13\cos^2\alpha = 1 + x^2$
$13(\sin^2\alpha + \cos^2\alpha) = 1 + x^2$
$13(1) = 1 + x^2$
$13 = 1 + x^2$
$x^2 = 12$
$x = \pm\sqrt{12}$
Multiply $2\sin\alpha + 3\cos\alpha = 1$ by 2 and $3\sin\alpha – 2\cos\alpha = x$ by 3:
$4\sin\alpha + 6\cos\alpha = 2$
$9\sin\alpha – 6\cos\alpha = 3x$
Adding these two equations gives:
$13\sin\alpha = 2 + 3x$
Multiply $2\sin\alpha + 3\cos\alpha = 1$ by 3 and $3\sin\alpha – 2\cos\alpha = x$ by 2:
$6\sin\alpha + 9\cos\alpha = 3$
$6\sin\alpha – 4\cos\alpha = 2x$
Subtracting the two equations gives:
$13\cos\alpha = 3 – 2x$
From $2\sin\alpha + 3\cos\alpha = 1$, we have $\sin\alpha = \frac{1 – 3\cos\alpha}{2}$.
So $x = 3(\frac{1 – 3\cos\alpha}{2}) – 2\cos\alpha = \frac{3}{2} – \frac{9}{2}\cos\alpha – 2\cos\alpha$
$x = \frac{3}{2} – \frac{13}{2}\cos\alpha$. Since $13\cos\alpha = 3 – 2x$, then $\cos\alpha = \frac{3 – 2x}{13}$.
$x = \frac{3}{2} – \frac{13}{2}(\frac{3-2x}{13}) = \frac{3}{2} – \frac{3-2x}{2} = \frac{3-3+2x}{2} = x$.
If we take the equations
$2\sin\alpha + 3\cos\alpha = 1$ and $3\sin\alpha – 2\cos\alpha = x$. Then by squaring and adding the two equations, we get
$(2\sin\alpha + 3\cos\alpha)^2 + (3\sin\alpha – 2\cos\alpha)^2 = 1^2 + x^2$
$4\sin^2\alpha + 12\sin\alpha\cos\alpha + 9\cos^2\alpha + 9\sin^2\alpha – 12\sin\alpha\cos\alpha + 4\cos^2\alpha = 1 + x^2$
$13(\sin^2\alpha + \cos^2\alpha) = 1 + x^2$
$13 = 1 + x^2$, so $x^2 = 12$, so $x = \pm 2\sqrt{3}$.
Consider $2(2\sin\alpha+3\cos\alpha) + 3(3\sin\alpha-2\cos\alpha) = 4\sin\alpha+6\cos\alpha+9\sin\alpha-6\cos\alpha = 13\sin\alpha$
$2(1) + 3x = 13\sin\alpha$
$3(2\sin\alpha+3\cos\alpha) – 2(3\sin\alpha-2\cos\alpha) = 6\sin\alpha+9\cos\alpha-6\sin\alpha+4\cos\alpha = 13\cos\alpha$
$3(1)-2x=13\cos\alpha$
$\sin^2\alpha+\cos^2\alpha = 1$
$(\frac{2+3x}{13})^2+(\frac{3-2x}{13})^2=1$
$4+12x+9x^2+9-12x+4x^2 = 169$
$13x^2+13=169$
$13x^2=156$
$x^2=12$
$x=\pm 2\sqrt{3}$
Since $2\sin\alpha+3\cos\alpha = 1$, $\sin\alpha>0$ and $\cos\alpha<0$. Then $3\sin\alpha - 2\cos\alpha>0$, so $x = 2\sqrt{3}$.

Answer: $2\sqrt{3}$

Ans: D

Solution:
Since $\tan 3B = \cot(B + 10^{\circ})$, we have $\tan 3B = \tan(90^{\circ} – (B + 10^{\circ}))$. Thus, $3B = 90^{\circ} – (B + 10^{\circ})$, which implies $3B = 90^{\circ} – B – 10^{\circ}$, so $4B = 80^{\circ}$, and $B = 20^{\circ}$. Since $3B = 60^{\circ}$ is an acute angle, this solution is valid.

Answer: $20^{\circ}$

Ans: B

Solution:
Since $\cos B = \frac{5}{13}$, we have a right triangle where the adjacent side is 5 and the hypotenuse is 13. Using the Pythagorean theorem, the opposite side is $\sqrt{13^2 – 5^2} = \sqrt{169-25} = \sqrt{144} = 12$.
Then, $\sin B = \frac{12}{13}$ and $\cot B = \frac{5}{12}$.
Answer: $\sin B = \frac{12}{13}, \cot B = \frac{5}{12}$

Ans: C

Solution:
$(\sec \theta – \tan \theta)^2 = \left(\frac{1}{\cos \theta} – \frac{\sin \theta}{\cos \theta}\right)^2 = \left(\frac{1 – \sin \theta}{\cos \theta}\right)^2 = \frac{(1 – \sin \theta)^2}{\cos^2 \theta} = \frac{(1 – \sin \theta)^2}{1 – \sin^2 \theta} = \frac{(1 – \sin \theta)^2}{(1 – \sin \theta)(1 + \sin \theta)} = \frac{1 – \sin \theta}{1 + \sin \theta}$
Answer: $\frac{1 – \sin \theta}{1 + \sin \theta}$

Ans: C

Solution:
$\log \sin 1^\circ + \log \sin 2^\circ + \log \sin 3^\circ + \dots + \log \sin 89^\circ = \log(\sin 1^\circ \sin 2^\circ \dots \sin 89^\circ)$
We know that $\sin x = \cos(90-x)$, so we can rewrite the product as:
$\sin 1^\circ \sin 2^\circ \dots \sin 44^\circ \sin 45^\circ \sin 46^\circ \dots \sin 89^\circ$
$= \sin 1^\circ \sin 2^\circ \dots \sin 44^\circ \cdot \frac{1}{\sqrt{2}} \cdot \cos 44^\circ \cos 43^\circ \dots \cos 1^\circ$
$= \frac{1}{\sqrt{2}} (\sin 1^\circ \cos 1^\circ) (\sin 2^\circ \cos 2^\circ) \dots (\sin 44^\circ \cos 44^\circ)$
$= \frac{1}{\sqrt{2}} \cdot \frac{1}{2^{44}}(\sin 2^\circ \sin 4^\circ \dots \sin 88^\circ)$
$= \frac{1}{\sqrt{2}} \frac{1}{2^{44}} (\sin 2^\circ \sin 4^\circ \dots \sin 88^\circ)$
Let $P = \sin 1^\circ \sin 2^\circ \dots \sin 89^\circ$
We have $\sin x = \sin (180-x)$ and $\sin (90-x) = \cos x$
$P = (\sin 1^\circ \sin 2^\circ \dots \sin 44^\circ) \sin 45^\circ (\sin 46^\circ \dots \sin 89^\circ)$
$P = \sin 1^\circ \sin 2^\circ \dots \sin 44^\circ \cdot \frac{1}{\sqrt{2}} \cdot \sin(90-44)^\circ \dots \sin(90-1)^\circ$
$P = \frac{1}{\sqrt{2}} \prod_{k=1}^{44} \sin k^\circ \cos k^\circ$
$P = \frac{1}{\sqrt{2}} \prod_{k=1}^{44} \frac{1}{2} \sin(2k^\circ)$
$P = \frac{1}{2^{44} \sqrt{2}} \sin 2^\circ \sin 4^\circ \dots \sin 88^\circ$
$P = \frac{1}{2^{44.5}}$
The result is $\log \frac{1}{2^{44.5}} = -44.5 \log 2$

Answer: $-44.5 \log 2$