Trigonometry Problems Class 10: 10 Solved Examples

Problem: Prove that $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

Solution:
Let's start with the Left Hand Side (LHS):

Hence, $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.

Problem: If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.

Solution:
Given $3 \cot A = 4$, which means $\cot A = \frac{4}{3}$.
Since $\tan A = \frac{1}{\cot A}$, we have $\tan A = \frac{3}{4}$.

Now, let's find $\sin A$ and $\cos A$. We can draw a right-angled triangle ABC, right-angled at B.
If $\cot A = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{AB}{BC} = \frac{4}{3}$.
Let $AB = 4k$ and $BC = 3k$.
By Pythagoras theorem, $AC^2 = AB^2 + BC^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2$.
So, $AC = \sqrt{25k^2} = 5k$.

Now, we can find $\sin A$ and $\cos A$:
$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5}$
$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5}$

Now, let's evaluate the LHS of the given equation:

Now, let's evaluate the RHS of the given equation:

Since LHS = RHS, the given statement is true.

Hence, $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ is true when $3 \cot A = 4$.

Problem: A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^\circ$. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $30^\circ$. Find the height of the tower and the width of the canal.

Solution:
Let AB be the height of the TV tower (h) and BC be the width of the canal (x).
Let C be the point on the other bank directly opposite the tower. The angle of elevation from C is $60^\circ$.
Let D be the point 20 m away from C on the line joining C to the foot of the tower B. So, $CD = 20$ m. The angle of elevation from D is $30^\circ$.

In right-angled $\triangle ABC$:
$\tan 60^\circ = \frac{AB}{BC}$
$\sqrt{3} = \frac{h}{x}$
$h = x\sqrt{3}$ ...(1)

In right-angled $\triangle ABD$:
$\tan 30^\circ = \frac{AB}{BD}$
$\frac{1}{\sqrt{3}} = \frac{h}{BC + CD}$
$\frac{1}{\sqrt{3}} = \frac{h}{x + 20}$ ...(2)

Substitute $h = x\sqrt{3}$ from (1) into (2):
$\frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 20}$
Cross-multiply:
$x + 20 = x\sqrt{3} \times \sqrt{3}$
$x + 20 = 3x$
$20 = 3x - x$
$20 = 2x$
$x = 10$ m

Now, substitute the value of x back into (1) to find h:
$h = x\sqrt{3} = 10\sqrt{3}$ m

Therefore, the height of the tower is $10\sqrt{3}$ m and the width of the canal is 10 m.

(A diagram showing a vertical tower AB, point C on one bank, point D 20m away from C on the same line, with angles of elevation $60^\circ$ from C and $30^\circ$ from D, would typically accompany this problem.)

Problem: If $x = a \sec \theta + b \tan \theta$ and $y = a \tan \theta + b \sec \theta$, prove that $x^2 - y^2 = a^2 - b^2$.

Solution:
Given:
$x = a \sec \theta + b \tan \theta$ ...(1)
$y = a \tan \theta + b \sec \theta$ ...(2)

We need to prove $x^2 - y^2 = a^2 - b^2$.
Let's find $x^2$:

Now, let's find $y^2$:

Now, subtract (4) from (3):

Notice that the $2ab \sec \theta \tan \theta$ terms cancel out: