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

Exercise 8.1

Step 1: Recall the definition of cosine in a right triangle.

\(\cos A = \dfrac{\text{Base}}{\text{Hypotenuse}}\)

Here, \(\cos A = \dfrac{4}{5}\).

This means Base = 4 units, Hypotenuse = 5 units.

Step 2: Find the third side (Perpendicular) using Pythagoras theorem.

\( \text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2 \)

\( 5^2 = 4^2 + \text{Perpendicular}^2 \)

\( 25 = 16 + \text{Perpendicular}^2 \)

\( \text{Perpendicular}^2 = 25 - 16 = 9 \)

\( \text{Perpendicular} = 3 \)

Step 3: Recall the definition of tangent.

\( \tan A = \dfrac{\text{Perpendicular}}{\text{Base}} \)

\( \tan A = \dfrac{3}{4} \)

Final Answer: \(\dfrac{3}{4}\) (Option B).

Step 1: Recall the definition of sine.

\(\sin A = \dfrac{\text{opposite side}}{\text{hypotenuse}}\).

We are given \(\sin A = \dfrac{1}{2}\).

Step 2: Think about the standard angles (30°, 45°, 60°) where sine values are commonly used.

We know: \(\sin 30^\circ = \dfrac{1}{2}\).

So, \(A = 30^\circ\).

Step 3: Now recall the formula for cotangent.

\(\cot A = \dfrac{\cos A}{\sin A}\).

Step 4: At \(A = 30^\circ\),

\(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\).

Step 5: Substitute these values:

\(\cot 30^\circ = \dfrac{\cos 30^\circ}{\sin 30^\circ} = \dfrac{\tfrac{\sqrt{3}}{2}}{\tfrac{1}{2}}\).

Step 6: Simplify the fraction:

\(\dfrac{\tfrac{\sqrt{3}}{2}}{\tfrac{1}{2}} = \sqrt{3}\).

Final Answer: \(\cot A = \sqrt{3}\).

Step 1: Write the given expression clearly.

\[ E = \csc(75^\circ + \theta) - \sec(15^\circ - \theta) - \tan(55^\circ + \theta) + \cot(35^\circ - \theta) \]

Step 2: Recall the trigonometric identities:

• \(\csc(90^\circ - x) = \sec(x)\)
• \(\tan(90^\circ - x) = \cot(x)\)

Step 3: Try to write angles so they fit the form \(90^\circ - x\).

Notice that:

• \(75^\circ + \theta = 90^\circ - (15^\circ - \theta)\)
• \(55^\circ + \theta = 90^\circ - (35^\circ - \theta)\)

Step 4: Apply the identities:

• \(\csc(75^\circ + \theta) = \csc[90^\circ - (15^\circ - \theta)] = \sec(15^\circ - \theta)\)
• \(\tan(55^\circ + \theta) = \tan[90^\circ - (35^\circ - \theta)] = \cot(35^\circ - \theta)\)

Step 5: Substitute these results back into \(E\):

\[ E = \sec(15^\circ - \theta) - \sec(15^\circ - \theta) - \cot(35^\circ - \theta) + \cot(35^\circ - \theta) \]

Step 6: Simplify. Each term cancels with its opposite:

\[ E = 0 \]

Final Answer: The value of the expression is 0.

We are given: \(\sin\theta = \dfrac{a}{b}\).

Step 1: Recall the Pythagorean identity of trigonometry:

\[ \sin^2\theta + \cos^2\theta = 1 \]

Step 2: Substitute the value of \(\sin\theta\):

\[ \left(\dfrac{a}{b}\right)^2 + \cos^2\theta = 1 \]

Step 3: Simplify the square:

\[ \dfrac{a^2}{b^2} + \cos^2\theta = 1 \]

Step 4: Move \(\dfrac{a^2}{b^2}\) to the right side:

\[ \cos^2\theta = 1 - \dfrac{a^2}{b^2} \]

Step 5: Write with a common denominator:

\[ \cos^2\theta = \dfrac{b^2}{b^2} - \dfrac{a^2}{b^2} = \dfrac{b^2 - a^2}{b^2} \]

Step 6: Take the square root on both sides:

\[ \cos\theta = \dfrac{\sqrt{b^2 - a^2}}{b} \]

Note: Since \(\theta\) is an acute angle (between \(0^\circ\) and \(90^\circ\)), cosine is positive. So we take only the positive root.

Final Answer: \(\cos\theta = \dfrac{\sqrt{b^2 - a^2}}{b}\) → Option C.

Step 1: We are given that \(\cos(\alpha + \beta) = 0\).

Step 2: Recall that \(\cos 90^\circ = 0\). So, for the cosine function to be zero, the angle must be \(90^\circ\) (or an odd multiple of it). For simplicity, take: \[ \alpha + \beta = 90^\circ \]

Step 3: From this, we can write: \[ \alpha = 90^\circ - \beta \]

Step 4: Now we want to reduce \(\sin(\alpha - \beta)\). Substitute the value of \(\alpha\): \[ \sin(\alpha - \beta) = \sin((90^\circ - \beta) - \beta) \]

Step 5: Simplify inside the bracket: \[ (90^\circ - \beta) - \beta = 90^\circ - 2\beta \]

Step 6: So, \[ \sin(\alpha - \beta) = \sin(90^\circ - 2\beta) \]

Step 7: Use the identity: \(\sin(90^\circ - \theta) = \cos\theta\). Here, \(\theta = 2\beta\). Therefore, \[ \sin(90^\circ - 2\beta) = \cos(2\beta) \]

Final Answer: \[ \sin(\alpha - \beta) = \cos 2\beta \]

Step 1: Write the given product:

\(P = \tan1^\circ \times \tan2^\circ \times \tan3^\circ \times \cdots \times \tan89^\circ\)

Step 2: Recall the trigonometric identity:

\(\tan(90^\circ - \theta) = \cot\theta = \dfrac{1}{\tan\theta}\)

Step 3: Pair the terms in the product. For example:

\(\tan1^\circ \times \tan89^\circ = \tan1^\circ \times \tan(90^\circ - 1^\circ) = \tan1^\circ \times \cot1^\circ = 1\)

\(\tan2^\circ \times \tan88^\circ = 1\)

\(\tan3^\circ \times \tan87^\circ = 1\)

And so on…

Step 4: Continue pairing up all terms this way. Each pair gives value = 1.

Step 5: The only angle left in the middle (which cannot be paired) is:

\(\tan45^\circ = 1\)

Step 6: Multiply everything:

\(P = (1 \times 1 \times 1 \times \cdots) \times 1 = 1\)

Final Answer: The value of the product is 1.

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

Step 2: In the question, we are given \(\cos 9\alpha = \sin \alpha\). Using the identity, replace \(\sin\alpha\) by \(\cos(90^\circ - \alpha)\).

So, \(\cos 9\alpha = \cos(90^\circ - \alpha)\).

Step 3: If two cos values are equal, their angles must also be equal (as long as both lie in the given range). Hence, \(9\alpha = 90^\circ - \alpha\).

Step 4: Solve for \(\alpha\):

\(9\alpha + \alpha = 90^\circ\)

\(10\alpha = 90^\circ\)

\(\alpha = 9^\circ\)

Step 5: Now we need \(\tan 5\alpha\).

Substitute \(\alpha = 9^\circ\):

\(\tan 5\alpha = \tan(5 \times 9^\circ) = \tan 45^\circ\).

Step 6: We know \(\tan 45^\circ = 1\).

Final Answer: \(\tan 5\alpha = 1\).

Step 1: In any triangle, the sum of the three angles is always \(180^\circ\).

So, \(A + B + C = 180^\circ\).

Step 2: The question says that the triangle is right-angled at \(C\). That means \(C = 90^\circ\).

Step 3: Substitute \(C = 90^\circ\) in the equation:

\(A + B + 90^\circ = 180^\circ\)

Step 4: Subtract \(90^\circ\) from both sides:

\(A + B = 90^\circ\)

Step 5: Now, we need to find \(\cos(A+B)\).

Since \(A+B = 90^\circ\), we get:

\(\cos(90^\circ) = 0\).

Final Answer: \(\cos(A+B) = 0\).

Step 1: We are given the equation:

\( \sin A + \sin^2 A = 1 \)

Step 2: Let us put \( \sin A = x \). Then the equation becomes:

\( x + x^2 = 1 \)

Step 3: Rearrange the equation:

\( x^2 + x - 1 = 0 \)

Step 4: Solve this quadratic equation. Using the quadratic formula: \( x = \dfrac{-1 \pm \sqrt{1 + 4}}{2} = \dfrac{-1 \pm \sqrt{5}}{2} \)

Step 5: Since \( \sin A \) must lie between \(-1\) and \(1\), only the value \( x = \dfrac{\sqrt{5} - 1}{2} \) is valid.

Step 6: Now, \( \sin^2 A = x^2 \). So, \( \cos^2 A = 1 - \sin^2 A = 1 - x^2 \).

Step 7: We need \( \cos^2 A + \cos^4 A \).

That is \( (1 - x^2) + (1 - x^2)^2 \).

Step 8: Expand:

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

Step 9: But from Step 3, we know \( x^2 = 1 - x \). Substitute this into the expression:

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

Step 10: Simplify:

\( 2 - 3 + 3x + (1 - 2x + x^2) = (0) + x^2 + x \).

Step 11: But from Step 2, \( x^2 + x = 1 \).

Final Answer: \( \cos^2 A + \cos^4 A = 1 \).

Step 1: We are told that \(\sin\alpha = \tfrac{1}{2}\).

From standard trigonometric values, \(\sin 30^\circ = \tfrac{1}{2}\).

So, \(\alpha = 30^\circ\) (taking the acute angle).

Step 2: We are also told that \(\cos\beta = \tfrac{1}{2}\).

From standard values, \(\cos 60^\circ = \tfrac{1}{2}\).

So, \(\beta = 60^\circ\) (taking the acute angle).

Step 3: Now we add the two angles.

\(\alpha + \beta = 30^\circ + 60^\circ = 90^\circ\).

Final Answer: \(90^\circ\).

Step 1: Look at the first fraction: \[ \dfrac{\sin^2 22^\circ + \sin^2 68^\circ}{\cos^2 22^\circ + \cos^2 68^\circ} \]

Notice that \(68^\circ = 90^\circ - 22^\circ\). So, \(\sin 68^\circ = \cos 22^\circ\).

Therefore, numerator becomes: \(\sin^2 22^\circ + (\cos 22^\circ)^2 = \sin^2 22^\circ + \cos^2 22^\circ = 1\).

Similarly, denominator is: \(\cos^2 22^\circ + (\sin 22^\circ)^2 = 1\).

So the fraction = \(\dfrac{1}{1} = 1\).

Step 2: Now consider the term \(\cos 63^\circ \sin 27^\circ\).

Notice \(27^\circ = 90^\circ - 63^\circ\). So, \(\sin 27^\circ = \cos 63^\circ\).

Therefore, \(\cos 63^\circ \cdot \sin 27^\circ = \cos 63^\circ \cdot \cos 63^\circ = \cos^2 63^\circ\).

Step 3: Look at the other term: \(\sin^2 63^\circ\).

We know the identity: \(\sin^2 \theta + \cos^2 \theta = 1\).

So, \(\sin^2 63^\circ + \cos^2 63^\circ = 1\).

Step 4: Put everything together:

Whole expression = fraction (which is 1) + \(\sin^2 63^\circ + \cos^2 63^\circ\) (which is also 1).

So total = \(1 + 1 = 2\).

Final Answer: \(2\) (Option B)

Step 1: We are given \(4 \tan\theta = 3\).

So, \(\tan\theta = \dfrac{3}{4}\).

Step 2: Recall that \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\).

This means we can imagine a right-angled triangle where:

• Opposite side (to angle \(\theta\)) = 3
• Adjacent side (to angle \(\theta\)) = 4

Step 3: Now find the hypotenuse using Pythagoras theorem:

\(\text{Hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5.\)

Step 4: From the triangle:

• \(\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}} = \dfrac{3}{5}\)
• \(\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}} = \dfrac{4}{5}\)

Step 5: Substitute these values into the given expression:

\(\dfrac{4\sin\theta - \cos\theta}{4\sin\theta + \cos\theta} = \dfrac{4\times \dfrac{3}{5} - \dfrac{4}{5}}{4\times \dfrac{3}{5} + \dfrac{4}{5}}\)

Step 6: Simplify numerator and denominator:

Numerator = \(\dfrac{12}{5} - \dfrac{4}{5} = \dfrac{8}{5}\)

Denominator = \(\dfrac{12}{5} + \dfrac{4}{5} = \dfrac{16}{5}\)

Step 7: Ratio = \(\dfrac{\tfrac{8}{5}}{\tfrac{16}{5}} = \dfrac{8}{16} = \dfrac{1}{2}\).

Final Answer: Option C (\(\dfrac{1}{2}\)).

Step 1: We are given that \(\sin\theta - \cos\theta = 0\).

This means \(\sin\theta = \cos\theta\).

Step 2: From trigonometry, we know that:

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

Step 3: Since \(\sin\theta = \cos\theta\), let each of them be equal to \(x\).

So, \(\sin\theta = x\) and \(\cos\theta = x\).

Step 4: Substitute into the identity:

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

\(2x^2 = 1\)

\(x^2 = \dfrac{1}{2}\)

\(x = \dfrac{1}{\sqrt{2}}\)

Step 5: Now calculate the required expression:

\(\sin^4\theta + \cos^4\theta = x^4 + x^4 = 2x^4\)

Step 6: Since \(x = \dfrac{1}{\sqrt{2}}\):

\(x^4 = \left(\dfrac{1}{\sqrt{2}}\right)^4 = \dfrac{1}{4}\)

So, \(2x^4 = 2 \times \dfrac{1}{4} = \dfrac{1}{2}\).

Final Answer: \(\sin^4\theta + \cos^4\theta = \dfrac{1}{2}\).

Let us simplify step by step:

Step 1: Recall the trigonometric expansion formulas:

• \(\sin(A+B) = \sin A \cos B + \cos A \sin B\)
• \(\cos(A-B) = \cos A \cos B + \sin A \sin B\)

Step 2: Apply these formulas one by one.

For \(\sin(45^\circ+\theta)\):

\(\sin(45^\circ+\theta) = \sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta\)

Step 3: We know that \(\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}}\).

So, \(\sin(45^\circ+\theta) = \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta)\).

Step 4: Now expand \(\cos(45^\circ-\theta)\):

\(\cos(45^\circ-\theta) = \cos 45^\circ \cos \theta + \sin 45^\circ \sin \theta\)

Again, \(\cos 45^\circ = \sin 45^\circ = \dfrac{1}{\sqrt{2}}\).

So, \(\cos(45^\circ-\theta) = \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta)\).

Step 5: Now subtract the two expressions:

\(\sin(45^\circ+\theta) - \cos(45^\circ-\theta) = \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta) - \dfrac{1}{\sqrt{2}}(\cos \theta + \sin \theta)\)

Step 6: Since both terms are exactly the same, their difference is:

\(0\).

Final Answer: Option B (0).

Step 1: Draw a right-angled triangle.

- The pole is the vertical side (opposite side) = \(6\,\text{m}\).

- The shadow is the horizontal side (adjacent side) = \(2\sqrt{3}\,\text{m}\).

- The angle of elevation of the Sun is \(\theta\) (the angle between the sunlight and the ground).

Step 2: Use the definition of tangent.

\( \tan\theta = \dfrac{\text{opposite side}}{\text{adjacent side}} \)

Here, opposite = height of pole = \(6\,\text{m}\), adjacent = shadow length = \(2\sqrt{3}\,\text{m}\).

Step 3: Substitute the values.

\( \tan\theta = \dfrac{6}{2\sqrt{3}} \)

Step 4: Simplify the fraction.

\( \dfrac{6}{2\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3} \)

Step 5: Recall the trigonometric ratio.

\( \tan 60^\circ = \sqrt{3} \)

Step 6: Therefore, \( \theta = 60^\circ \).

Final Answer: The Sun’s elevation is \(60^\circ\).