NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 8: Introduction to Trignometry & its Equation

Exercise 8.2

Step 1: Recall the relation between cotangent and tangent:

\(\cot \theta = \tan(90^\circ - \theta)\).

Step 2: Put \(\theta = 43^\circ\).

So, \(\cot 43^\circ = \tan(90^\circ - 43^\circ)\).

Step 3: Simplify inside the bracket:

\(90^\circ - 43^\circ = 47^\circ\).

So, \(\cot 43^\circ = \tan 47^\circ\).

Step 4: Now substitute this result into the given expression:

\(\dfrac{\tan 47^\circ}{\cot 43^\circ} = \dfrac{\tan 47^\circ}{\tan 47^\circ}\).

Step 5: Anything divided by itself is 1 (except zero).

So, \(\dfrac{\tan 47^\circ}{\tan 47^\circ} = 1\).

Final Answer: The statement is True.

Step 1: Recall the trigonometric identity: \(\sin(90^\circ - \theta) = \cos \theta\).

Step 2: Here, we have \(\sin 67^\circ\). Notice that \(67^\circ = 90^\circ - 23^\circ\).

Step 3: So, \(\sin 67^\circ = \sin(90^\circ - 23^\circ) = \cos 23^\circ\).

Step 4: Substitute this value into the expression: \(\cos^2 23^\circ - \sin^2 67^\circ = \cos^2 23^\circ - (\cos 23^\circ)^2\).

Step 5: Simplify: \(\cos^2 23^\circ - \cos^2 23^\circ = 0\).

Step 6: The result is \(0\). Since 0 is neither positive nor negative, the statement “the value is positive” is False.

Let us carefully check step by step:

Step 1: Recall the meaning of sine and cosine values.
- For any angle \(\theta\) (measured in degrees here), \(\sin \theta\) and \(\cos \theta\) are numbers between 0 and 1 (in SI system, they are pure ratios, so they do not have any units).

Step 2: Compare sine and cosine in the range \(45^\circ \leq \theta \leq 90^\circ\).
- At \(45^\circ\), both are equal: \(\sin 45^\circ = \cos 45^\circ = 0.707\).
- For angles larger than \(45^\circ\), the sine value keeps increasing towards 1, while the cosine value keeps decreasing towards 0.

Step 3: At \(\theta = 80^\circ\):
- \(\sin 80^\circ \approx 0.985\)
- \(\cos 80^\circ \approx 0.173\)

Step 4: Subtract the two values:
\[ \sin 80^\circ - \cos 80^\circ = 0.985 - 0.173 = 0.812 \]

Step 5: The result (0.812) is positive, not negative.

Final Conclusion: The given statement is False. The expression \(\sin 80^\circ - \cos 80^\circ\) is positive.

Let us check the statement step by step:

Step 1: Recall the identity:

\(\sin^2\theta + \cos^2\theta = 1\).

So, \(1 - \cos^2\theta = \sin^2\theta\).

Step 2: Now take the square root:

\(\sqrt{1 - \cos^2\theta} = \sqrt{\sin^2\theta} = |\sin\theta|\).

(We use absolute value because square root always gives a non-negative result.)

Step 3: Write the Left Hand Side (LHS):

\(LHS = \sqrt{1 - \cos^2\theta} \cdot \sec^2\theta\).

Substitute: \(LHS = |\sin\theta| \cdot \sec^2\theta\).

Step 4: Replace \(\sec\theta\) with \(1/\cos\theta\):

\(\sec^2\theta = \dfrac{1}{\cos^2\theta}\).

So, \(LHS = \dfrac{|\sin\theta|}{\cos^2\theta}\).

Step 5: Write the Right Hand Side (RHS):

\(RHS = \tan\theta = \dfrac{\sin\theta}{\cos\theta}\).

Step 6: Compare LHS and RHS:

• LHS = \(\dfrac{|\sin\theta|}{\cos^2\theta}\)
• RHS = \(\dfrac{\sin\theta}{\cos\theta}\)

They are not the same in general.

Step 7: Check with an example:

Take \(\theta = 45^\circ\).

\(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\), \(\cos 45^\circ = \dfrac{\sqrt{2}}{2}\).

LHS = \(\dfrac{\sqrt{2}/2}{(\sqrt{2}/2)^2} = \dfrac{\sqrt{2}/2}{1/2} = \sqrt{2}\).

RHS = \(\dfrac{\sqrt{2}/2}{\sqrt{2}/2} = 1\).

Conclusion: LHS \(\neq\) RHS. Hence, the statement is False.

Step 1: Let \(x = \cos A\). The given condition is:

\(x + x^2 = 1\).

Step 2: Rearrange to find \(x^2\):

\(x^2 = 1 - x\).

Step 3: Recall that \(\sin^2 A = 1 - \cos^2 A\).

So, \(\sin^2 A = 1 - x^2\).

Step 4: Now calculate \(\sin^2 A + \sin^4 A\):

\(\sin^2 A + \sin^4 A = (1 - x^2) + (1 - x^2)^2\).

Step 5: Expand the square:

\((1 - x^2)^2 = 1 - 2x^2 + x^4\).

So, \(\sin^2 A + \sin^4 A = (1 - x^2) + (1 - 2x^2 + x^4)\).

Step 6: Simplify:

\(\sin^2 A + \sin^4 A = 2 - 3x^2 + x^4\).

Step 7: Replace \(x^2\) using Step 2: \(x^2 = 1 - x\).

So, \(x^4 = (x^2)^2 = (1 - x)^2 = 1 - 2x + x^2\).

But \(x^2 = 1 - x\), so:

\(x^4 = 1 - 2x + (1 - x) = 2 - 3x\).

Step 8: Substitute this back:

\(2 - 3x^2 + x^4 = 2 - 3(1 - x) + (2 - 3x)\).

Step 9: Simplify step by step:

= \(2 - 3 + 3x + 2 - 3x\)

= \(1\).

Final Result: Therefore, \(\sin^2 A + \sin^4 A = 1\). The statement is True.

Step 1: Expand the left-hand side (LHS).

LHS = \((\tan\theta + 2)(2\tan\theta + 1)\)

Use distributive property: \(a(b+c) = ab + ac\).

= \(\tan\theta(2\tan\theta + 1) + 2(2\tan\theta + 1)\)

= \(2\tan^2\theta + \tan\theta + 4\tan\theta + 2\)

= \(2\tan^2\theta + 5\tan\theta + 2\).

Step 2: Compare with the right-hand side (RHS).

RHS = \(5\tan\theta + \sec^2\theta\).

Step 3: Simplify the LHS using identity.

We know the trigonometric identity: \(1 + \tan^2\theta = \sec^2\theta\).

So, \(2\tan^2\theta + 2 = 2(\tan^2\theta + 1) = 2\sec^2\theta\).

Therefore, LHS = \(5\tan\theta + 2\sec^2\theta\).

Step 4: Final comparison.

LHS = \(5\tan\theta + 2\sec^2\theta\)

RHS = \(5\tan\theta + \sec^2\theta\)

Since \(2\sec^2\theta \neq \sec^2\theta\), the two sides are not equal.

Therefore, the given statement is False.

Step 1: Suppose the height of the tower is \(h\) metres (a fixed value). Let the length of the shadow on the ground be \(d\) metres.

Step 2: In a right-angled triangle, the angle of elevation \(\theta\) of the sun is given by:

\[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{h}{d} \]

Step 3: Since the tower’s height \(h\) is constant, the only changing quantity is the shadow length \(d\).

Step 4: If the shadow becomes longer (\(d\) increases in metres), the fraction \(\dfrac{h}{d}\) becomes smaller.

Step 5: When \(\tan \theta\) decreases, the angle \(\theta\) (angle of elevation of the sun) also decreases.

Step 6: Therefore, as the shadow length increases, the angle of elevation of the sun decreases, not increases.

Step 1: The man is standing at a point 3 m above the surface of the lake (height measured in SI units).

Step 2: The cloud is above the lake. Its reflection will appear the same distance below the surface of the lake, because water acts like a horizontal mirror.

Step 3: From the man’s eye, one line of sight goes upward to the real cloud (this is the angle of elevation), and another line of sight goes downward to the reflection (this is the angle of depression).

Step 4: The vertical distance from the man’s eye to the cloud is the same as the vertical distance from his eye to the reflection. This is because the reflection is located symmetrically below the water surface.

Step 5: These equal vertical distances form two identical right-angled triangles — one with the cloud and one with the reflection.

Step 6: Since the two triangles are congruent, the angle of elevation (to the cloud) is equal to the angle of depression (to its reflection).

Therefore, the statement is True.

Step 1: We are told that \(a > 0\) and \(a \neq 1\).

Step 2: From the AM–GM inequality (Arithmetic Mean – Geometric Mean), we know that for any positive number \(a\):

\[ a + \frac{1}{a} \geq 2, \]

and equality (i.e., exactly equal to 2) happens only when \(a = 1\).

Step 3: Since the problem says \(a \neq 1\), this means:

\[ a + \frac{1}{a} > 2. \]

Step 4: Now, let’s check the possible values of \(2\sin\theta\).
We know that \(-1 \leq \sin\theta \leq 1\).
So multiplying the whole inequality by 2 gives:

\[ -2 \leq 2\sin\theta \leq 2. \]

Step 5: This means that \(2\sin\theta\) can never be greater than 2. Its maximum possible value is exactly 2, and minimum is -2.

Step 6: But from Step 3, \(a + \dfrac{1}{a} > 2\).
So \(a + \dfrac{1}{a}\) is always bigger than the maximum possible value of \(2\sin\theta\).

Final Step: Therefore, \(2\sin\theta\) can never equal \(a + \dfrac{1}{a}\) (except at \(a = 1\), which is not allowed).
Hence, the statement is False.

Step 1: Recall that for any angle \(\theta\), the value of cosine is always between -1 and 1.

That is: \(-1 \leq \cos\theta \leq 1\).

Step 2: Now look at the right-hand side: \(\dfrac{a^2+b^2}{2ab}\).

Step 3: Use the AM–GM inequality:
For any positive numbers \(a\) and \(b\),
\(\dfrac{a^2+b^2}{2} \geq ab\),
and equality holds only if \(a=b\).

Step 4: Divide both sides by \(ab\gt0\):
\(\dfrac{a^2+b^2}{2ab} \geq 1\),
with equality only if \(a=b\).

Step 5: But in our question, \(a\) and \(b\) are distinct (not equal).
So, \(\dfrac{a^2+b^2}{2ab} \gt 1\).

Step 6: This means the RHS is always greater than 1.
But cosine values cannot be greater than 1.
Therefore, the given relation cannot be true.

Final Conclusion: The statement is False.

Step 1: Let the original height of the tower be \(h\) metres, and the horizontal distance of the observer from the base of the tower be \(d\) metres.

Step 2: By definition of tangent,

\[ \tan 30^\circ = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{h}{d} \]

So, \( \dfrac{h}{d} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}. \)

Step 3: Now, suppose the height of the tower is doubled. The new height = \(2h\) metres.

Let the new angle of elevation be \(\theta'\).

Again, using tangent:

\[ \tan \theta' = \frac{2h}{d} \]

Step 4: From Step 2, we know \( \dfrac{h}{d} = \tan 30^\circ. \)

So, \( \tan \theta' = 2 \times \tan 30^\circ. \)

\[ \tan \theta' = 2 \times \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \approx 1.1547 \]

Step 5: Now find \(\theta'\):

\[ \theta' = \tan^{-1}(1.1547) \approx 49.4^\circ \]

Step 6: The new angle of elevation is about \(49.4^\circ\), not \(60^\circ\).
Therefore, the angle does not double when the height of the tower is doubled.

Let the original height of the tower be \(h\) metres, and the distance of the point of observation from the foot of the tower be \(d\) metres.

The angle of elevation \(\theta\) is given by the formula:

\[ \tan \theta = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{h}{d} \]

Now, if both the height and the distance are increased by 10%:

• New height = \(h + 10\%\,\text{of}\,h = 1.1h\) metres
• New distance = \(d + 10\%\,\text{of}\,d = 1.1d\) metres

The new angle of elevation is \(\theta'\), and:

\[ \tan \theta' = \frac{1.1h}{1.1d} \]

Simplify the fraction:

\[ \tan \theta' = \frac{1.1}{1.1} \times \frac{h}{d} = \frac{h}{d} \]

This means:

\[ \tan \theta' = \tan \theta \]

So, \(\theta' = \theta\).

Therefore, the angle of elevation remains unchanged even if both the height and the distance are increased by the same percentage.