Introduction to Trigonometry — NCERT Solutions, AI Tutor & Practice

In right triangle $ABC$, right-angled at $B$, the hypotenuse is $AC$.

Using Pythagoras theorem: $AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625$ So $AC = 25$ cm.

For angle $A$:

• Side opposite to $A$ = $BC = 7$
• Side adjacent to $A$ = $AB = 24$
• Hypotenuse = $AC = 25$

Therefore: $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}$ $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24}{25}$

Given $\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{4}$.

Let the opposite side $= 3k$ and adjacent side $= 4k$ for some positive $k$.

By Pythagoras theorem, hypotenuse: $= \sqrt{(3k)^2 + (4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k$

Therefore: $\sin A = \frac{3k}{5k} = \frac{3}{5}$ $\cos A = \frac{4k}{5k} = \frac{4}{5}$

Given $15\cot A = 8$, so $\cot A = \dfrac{8}{15} = \dfrac{\text{adjacent}}{\text{opposite}}$.

Let adjacent $= 8k$ and opposite $= 15k$.

Hypotenuse $= \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$.

Therefore: $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15k}{17k} = \frac{15}{17}$ $\sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{17k}{8k} = \frac{17}{8}$

Given $\sin \theta = \dfrac{5}{13} = \dfrac{\text{opposite}}{\text{hypotenuse}}$.

Let opposite $= 5k$, hypotenuse $= 13k$.

Adjacent $= \sqrt{(13k)^2 - (5k)^2} = \sqrt{169k^2 - 25k^2} = \sqrt{144k^2} = 12k$.

So $\cos\theta = \dfrac{12}{13}$.

Now substitute: $\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{\frac{12}{13} - \frac{5}{13}}{\frac{12}{13} + \frac{5}{13}} = \frac{\frac{7}{13}}{\frac{17}{13}} = \frac{7}{17}$

Let $QR = x$ cm. Then $PR = (25 - x)$ cm and $PQ = 5$ cm.

By Pythagoras theorem ($PR$ is the hypotenuse): $PR^2 = PQ^2 + QR^2$ $(25 - x)^2 = 5^2 + x^2$ $625 - 50x + x^2 = 25 + x^2$ $625 - 50x = 25$ $50x = 600 \implies x = 12$

So $QR = 12$ cm and $PR = 25 - 12 = 13$ cm.

For angle $P$: opposite $= QR = 12$, adjacent $= PQ = 5$, hypotenuse $= PR = 13$. $\sin P = \frac{12}{13}, \quad \cos P = \frac{5}{13}, \quad \tan P = \frac{12}{5}$

Consider a right triangle $ABC$ right-angled at $C$, with $\angle A$ and $\angle B$ acute.

By definition: $\cos A = \frac{AC}{AB}, \qquad \cos B = \frac{BC}{AB}$

Given $\cos A = \cos B$: $\frac{AC}{AB} = \frac{BC}{AB}$

Since the denominators are equal: $AC = BC$

In a triangle, angles opposite equal sides are equal. Here $AC = BC$, so the angles opposite them are equal: $\angle B = \angle A$

Hence $\angle A = \angle B$.

Using standard values: $\sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}, \quad \cos 60^\circ = \frac{1}{2}$

Substitute: $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\cdot\frac{1}{2}$ $= \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$

(This is in fact $\sin(60^\circ + 30^\circ) = \sin 90^\circ = 1$.)

Standard values: $\tan 45^\circ = 1, \quad \cot 45^\circ = 1, \quad \cos 30^\circ = \frac{\sqrt3}{2}, \quad \sin 60^\circ = \frac{\sqrt3}{2}$

Numerator: $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ = 2(1)^2 + \left(\frac{\sqrt3}{2}\right)^2 - \left(\frac{\sqrt3}{2}\right)^2$ $= 2 + \frac{3}{4} - \frac{3}{4} = 2$

Denominator: $\cot^2 45^\circ = 1^2 = 1$

Therefore the value $= \dfrac{2}{1} = 2$.

We know: $\tan 60^\circ = \sqrt3 \implies A + B = 60^\circ \quad (1)$ $\tan 30^\circ = \frac{1}{\sqrt3} \implies A - B = 30^\circ \quad (2)$

Adding (1) and (2): $2A = 90^\circ \implies A = 45^\circ$

Substituting into (1): $45^\circ + B = 60^\circ \implies B = 15^\circ$

Hence $A = 45^\circ$ and $B = 15^\circ$.

Expand the left-hand side (LHS): $(\sin\theta + \csc\theta)^2 = \sin^2\theta + 2\sin\theta\csc\theta + \csc^2\theta$ Since $\sin\theta\csc\theta = 1$: $= \sin^2\theta + 2 + \csc^2\theta$

Similarly: $(\cos\theta + \sec\theta)^2 = \cos^2\theta + 2 + \sec^2\theta$

Adding: $\text{LHS} = (\sin^2\theta + \cos^2\theta) + 4 + \csc^2\theta + \sec^2\theta$ $= 1 + 4 + \csc^2\theta + \sec^2\theta$

Using identities $\csc^2\theta = 1 + \cot^2\theta$ and $\sec^2\theta = 1 + \tan^2\theta$: $= 5 + (1 + \cot^2\theta) + (1 + \tan^2\theta)$ $= 7 + \tan^2\theta + \cot^2\theta = \text{RHS}$

Hence proved.

Take the common denominator on the LHS: $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}$

Expand the numerator: $\cos^2 A + 1 + 2\sin A + \sin^2 A$

Using $\sin^2 A + \cos^2 A = 1$: $= 1 + 1 + 2\sin A = 2 + 2\sin A = 2(1 + \sin A)$

So: $\text{LHS} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} = 2\sec A = \text{RHS}$

Hence proved.

Given $\sec\theta + \tan\theta = p$.

Using the identity $\sec^2\theta - \tan^2\theta = 1$: $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$ $p(\sec\theta - \tan\theta) = 1 \implies \sec\theta - \tan\theta = \frac{1}{p}$

Now: $p^2 - 1 = p^2 - 1, \quad \text{and note}$ $p - \frac{1}{p} = (\sec\theta + \tan\theta) - (\sec\theta - \tan\theta) = 2\tan\theta$ $p + \frac{1}{p} = (\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = 2\sec\theta$

Compute: $\frac{p^2 - 1}{p^2 + 1} = \frac{p - \frac{1}{p}}{p + \frac{1}{p}} \quad (\text{dividing numerator and denominator by } p)$ $= \frac{2\tan\theta}{2\sec\theta} = \frac{\tan\theta}{\sec\theta} = \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}} = \sin\theta$

Hence proved.