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

Exercise 8.3

Step 1: We start with the left-hand side (LHS):

\(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}\)

Step 2: To add these fractions, we take the common denominator, which is \(\sin\theta(1+\cos\theta)\).

Step 3: Write the fractions with this denominator:

\(\dfrac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}\)

Step 4: Expand the numerator:

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

Step 5: Recall the Pythagoras identity: \(\sin^2\theta + \cos^2\theta = 1\).

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

Step 6: Replace in the numerator:

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

Step 7: Factorize:

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

Step 8: Now the fraction becomes:

\(\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}\)

Step 9: Cancel \((1+\cos\theta)\) from numerator and denominator:

\(\dfrac{2}{\sin\theta}\)

Step 10: Recall that \(\csc\theta = \dfrac{1}{\sin\theta}\).

Final Step: Therefore,

\(\dfrac{2}{\sin\theta} = 2\csc\theta\).

Hence proved.

Step 1: Start with the Left Hand Side (LHS):

\(\displaystyle \dfrac{\tan A}{1+\sec A} - \dfrac{\tan A}{1-\sec A}\)

Step 2: Take a common denominator.

The denominators are \((1+\sec A)\) and \((1-\sec A)\). Their common denominator is \((1+\sec A)(1-\sec A)\).

Step 3: Write with the common denominator:

\(\displaystyle \frac{\tan A(1-\sec A) - \tan A(1+\sec A)}{(1+\sec A)(1-\sec A)}\)

Step 4: Expand the numerator:

\(= \tan A(1-\sec A) - \tan A(1+\sec A)\)

\(= \tan A - \tan A\sec A - \tan A - \tan A\sec A\)

\(= (\tan A - \tan A) - (\tan A\sec A + \tan A\sec A)\)

\(= 0 - 2\tan A\sec A\)

\(= -2\tan A\sec A\)

Step 5: Simplify the denominator:

\((1+\sec A)(1-\sec A) = 1 - \sec^2 A\)

Step 6: So now we have:

\(\displaystyle \frac{-2\tan A\sec A}{1 - \sec^2 A}\)

Step 7: Recall the trigonometric identity:

\(1 + \tan^2 A = \sec^2 A\)

So, \(1 - \sec^2 A = -\tan^2 A\).

Step 8: Substitute this into the denominator:

\(\displaystyle \frac{-2\tan A\sec A}{-\tan^2 A} = \frac{2\tan A\sec A}{\tan^2 A}\)

Step 9: Split the fraction:

\(= 2 \cdot \frac{\sec A}{\tan A}\)

Step 10: Write \(\sec A = \frac{1}{\cos A}\) and \(\tan A = \frac{\sin A}{\cos A}\):

\(\frac{\sec A}{\tan A} = \frac{\dfrac{1}{\cos A}}{\dfrac{\sin A}{\cos A}} = \frac{1}{\sin A}\)

Step 11: So we get:

\(2 \cdot \frac{1}{\sin A} = 2\csc A\)

Final Result: LHS = RHS = \(2\csc A\). Hence proved ✅

Step 1: Recall the meaning of \(\tan A\). It is defined as:

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

Step 2: Here, \(\tan A = \dfrac{3}{4}\). This means the length of the opposite side = 3 units, and the length of the adjacent side = 4 units.

Step 3: To find the hypotenuse of the right triangle, use the Pythagoras theorem:

\(\text{Hypotenuse} = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2}\).

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

Step 4: Now we can find sine and cosine:

• \(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{3}{5}\)
• \(\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{4}{5}\)

Step 5: Multiply sine and cosine:

\(\sin A \cdot \cos A = \dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{12}{25}\).

Final Answer: \(\sin A \cos A = \dfrac{12}{25}\).

Step 1: Write down what we need to prove.

We need to show that:

\[(\sin\alpha + \cos\alpha)(\tan\alpha + \cot\alpha) = \sec\alpha + \csc\alpha\]

Step 2: Expand the brackets on the left-hand side.

\((\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha)\)

= \(\sin\alpha\tan\alpha + \sin\alpha\cot\alpha + \cos\alpha\tan\alpha + \cos\alpha\cot\alpha\)

Step 3: Write tan and cot in terms of sin and cos.

\(\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}, \; \cot\alpha = \dfrac{\cos\alpha}{\sin\alpha}\)

Step 4: Substitute these values.

= \(\sin\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} + \sin\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha} + \cos\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} + \cos\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha}\)

Step 5: Simplify each term carefully.

• \(\sin\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\sin^2\alpha}{\cos\alpha}\)
• \(\sin\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha} = \cos\alpha\)
• \(\cos\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} = \sin\alpha\)
• \(\cos\alpha \cdot \dfrac{\cos\alpha}{\sin\alpha} = \dfrac{\cos^2\alpha}{\sin\alpha}\)

So the expression becomes:

\(= \dfrac{\sin^2\alpha}{\cos\alpha} + \cos\alpha + \sin\alpha + \dfrac{\cos^2\alpha}{\sin\alpha}\)

Step 6: Rearrange terms to group fractions together.

\(= \left(\dfrac{\sin^2\alpha}{\cos\alpha} + \cos\alpha\right) + \left(\dfrac{\cos^2\alpha}{\sin\alpha} + \sin\alpha\right)\)

Step 7: Simplify each group.

• \(\dfrac{\sin^2\alpha}{\cos\alpha} + \cos\alpha = \dfrac{\sin^2\alpha + \cos^2\alpha}{\cos\alpha} = \dfrac{1}{\cos\alpha} = \sec\alpha\)
• \(\dfrac{\cos^2\alpha}{\sin\alpha} + \sin\alpha = \dfrac{\cos^2\alpha + \sin^2\alpha}{\sin\alpha} = \dfrac{1}{\sin\alpha} = \csc\alpha\)

Step 8: Combine results.

= \(\sec\alpha + \csc\alpha\)

Therefore proved.

Step 1: Write down what is given.

We need to prove that

\[(\sqrt{3}+1)(3-\cot 30^\circ) = \tan^3 60^\circ - 2\sin 60^\circ.\]

Step 2: Recall trigonometric values.

• \(\cot 30^\circ = \sqrt{3}\)
• \(\tan 60^\circ = \sqrt{3}\)
• \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\)

Step 3: Simplify the Left-Hand Side (LHS).

LHS = \((\sqrt{3} + 1)(3 - \cot 30^\circ)\)

Substitute \(\cot 30^\circ = \sqrt{3}\):

LHS = \((\sqrt{3} + 1)(3 - \sqrt{3})\)

Now expand using distributive law:

= \( (\sqrt{3} \times 3) + (1 \times 3) - (\sqrt{3} \times \sqrt{3}) - (1 \times \sqrt{3}) \)

= \( 3\sqrt{3} + 3 - 3 - \sqrt{3} \)

= \( (3\sqrt{3} - \sqrt{3}) + (3 - 3) \)

= \( 2\sqrt{3} \)

So, LHS = \(2\sqrt{3}\).

Step 4: Simplify the Right-Hand Side (RHS).

RHS = \( \tan^3 60^\circ - 2\sin 60^\circ \)

Substitute values:

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

\( \sin 60^\circ = \dfrac{\sqrt{3}}{2} \Rightarrow 2\sin 60^\circ = 2 \times \dfrac{\sqrt{3}}{2} = \sqrt{3} \)

So RHS = \( 3\sqrt{3} - \sqrt{3} = 2\sqrt{3} \)

Step 5: Compare both sides.

LHS = \(2\sqrt{3}\), RHS = \(2\sqrt{3}\)

Since both are equal, the given equation is proved.

Step 1: Recall the trigonometric identities.

• \(\cot\alpha = \dfrac{\cos\alpha}{\sin\alpha} \implies \cot^2\alpha = \dfrac{\cos^2\alpha}{\sin^2\alpha}\)
• \(\csc\alpha = \dfrac{1}{\sin\alpha}\)

Step 2: Start with the left-hand side (LHS).

\[ LHS = 1 + \dfrac{\cot^2\alpha}{1 + \csc\alpha} \]

Step 3: Replace \(\cot^2\alpha\) and \(\csc\alpha\) with their \(\sin\alpha\) and \(\cos\alpha\) forms.

\[ LHS = 1 + \dfrac{\dfrac{\cos^2\alpha}{\sin^2\alpha}}{1 + \dfrac{1}{\sin\alpha}} \]

Step 4: Simplify the denominator \(1 + \dfrac{1}{\sin\alpha}\).

\[ 1 + \dfrac{1}{\sin\alpha} = \dfrac{\sin\alpha + 1}{\sin\alpha} \]

Step 5: Put this back into the fraction.

\[ LHS = 1 + \dfrac{\cos^2\alpha}{\sin^2\alpha} \times \dfrac{\sin\alpha}{\sin\alpha + 1} \]

Step 6: Simplify the multiplication.

\[ LHS = 1 + \dfrac{\cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]

Step 7: Write both terms of LHS with a common denominator.

\[ LHS = \dfrac{\sin\alpha(\sin\alpha + 1)}{\sin\alpha(\sin\alpha + 1)} + \dfrac{\cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]

\[ LHS = \dfrac{\sin^2\alpha + \sin\alpha + \cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]

Step 8: Use the Pythagoras identity \(\sin^2\alpha + \cos^2\alpha = 1\).

\[ LHS = \dfrac{1 + \sin\alpha}{\sin\alpha(\sin\alpha + 1)} \]

Step 9: Cancel \((1+\sin\alpha)\) from numerator and denominator.

\[ LHS = \dfrac{1}{\sin\alpha} \]

Step 10: Recall that \(\dfrac{1}{\sin\alpha} = \csc\alpha\).

Therefore, LHS = RHS = \(\csc\alpha\). ✔

Step 1: Recall the co-function identities:

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

Step 2: Write the Left-Hand Side (LHS):

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

Step 3: Express \(\tan\theta\) and \(\cot\theta\) in terms of \(\sin\theta\) and \(\cos\theta\):

• \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
• \(\cot\theta = \dfrac{\cos\theta}{\sin\theta}\)

Step 4: Add them together:

\(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta}\)

Step 5: Use the Pythagorean identity:

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

So, LHS = \(\dfrac{1}{\sin\theta \cos\theta}\)

Step 6: Now take the Right-Hand Side (RHS):

\(\sec\theta \cdot \sec(90^\circ - \theta) = \sec\theta \cdot \csc\theta\)

Step 7: Write secant and cosecant in terms of sine and cosine:

• \(\sec\theta = \dfrac{1}{\cos\theta}\)
• \(\csc\theta = \dfrac{1}{\sin\theta}\)

Step 8: Multiply them:

\(\dfrac{1}{\cos\theta} \times \dfrac{1}{\sin\theta} = \dfrac{1}{\sin\theta \cos\theta}\)

Step 9: Compare LHS and RHS:

LHS = RHS = \(\dfrac{1}{\sin\theta \cos\theta}\)

Final Result: The given identity is proved.

Step 1: Draw a right triangle to represent the situation.

• The vertical pole is the opposite side of the right triangle.
• The shadow on the ground is the adjacent side.
• The angle of elevation of the Sun is the angle \(\theta\) formed between the line from the top of the pole to the tip of the shadow and the ground.

Step 2: Write down the given information.

• Height of pole = \(h\) metres
• Length of shadow = \(\sqrt{3}\,h\) metres

Step 3: Recall the trigonometric ratio for tangent:

\[ \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \]

Step 4: Substitute the values.

\[ \tan(\theta) = \frac{h}{\sqrt{3}h} \]

Step 5: Simplify.

\[ \tan(\theta) = \frac{1}{\sqrt{3}} \]

Step 6: Recall the standard trigonometric value.

\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

Step 7: Therefore, \(\theta = 30^\circ\).

Step 1: We are given: \(\sqrt{3}\,\tan\theta = 1\).

Divide both sides by \(\sqrt{3}\):

\(\tan\theta = \dfrac{1}{\sqrt{3}}\).

Step 2: Recall the standard values of tan:

• \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\)
• \(\tan 45^\circ = 1\)
• \(\tan 60^\circ = \sqrt{3}\)

So, \(\theta = 30^\circ\) (since we consider an acute angle).

Step 3: Now we need to calculate \(\sin^2\theta - \cos^2\theta\).

Substitute \(\theta = 30^\circ\):

\(\sin^2 30^\circ - \cos^2 30^\circ\).

Step 4: Use the standard values of sine and cosine:

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

Step 5: Square them:

\(\sin^2 30^\circ = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\)

\(\cos^2 30^\circ = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}\)

Step 6: Subtract:

\(\sin^2 30^\circ - \cos^2 30^\circ = \dfrac{1}{4} - \dfrac{3}{4} = -\dfrac{2}{4} = -\dfrac{1}{2}.\)

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

Step 1: Draw the situation in your mind.

• The ladder is \(15\,\text{m}\) long. This is the hypotenuse of a right triangle.
• The wall is vertical, so the height of the wall will be the vertical side of the triangle.
• The ladder makes an angle of \(60^\circ\) with the wall.

Step 2: Identify which trigonometric ratio to use.

• We know the hypotenuse (ladder length = \(15\,\text{m}\)).
• We want to find the side adjacent to the angle with the wall (the vertical height).
• So we use the cosine function: \(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}\).

Step 3: Apply the formula.

\[ \cos 60^\circ = \frac{\text{Height of wall}}{15} \]

Step 4: Substitute the value of \(\cos 60^\circ\).

\[ \frac{1}{2} = \frac{\text{Height of wall}}{15} \]

Step 5: Solve for the height.

\[ \text{Height of wall} = 15 \times \frac{1}{2} = 7.5 \]

Step 6: Write the final answer with unit (SI).

\[ \text{Height of wall} = 7.5\,\text{m} \]

Step 1: Recall the trigonometric identity:

\[ 1 + \tan^2\theta = \sec^2\theta. \]

So we can replace \((1 + \tan^2\theta)\) with \(\sec^2\theta\).

Step 2: Look at the terms \((1 - \sin\theta)(1 + \sin\theta)\). This is in the form of \((a - b)(a + b) = a^2 - b^2.\)

Here, \(a = 1\) and \(b = \sin\theta\). So, \((1 - \sin\theta)(1 + \sin\theta) = 1^2 - (\sin\theta)^2 = 1 - \sin^2\theta.\)

Step 3: Another identity is:

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

Rearranging gives: \(1 - \sin^2\theta = \cos^2\theta.\)

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

Step 4: Now substitute back into the expression:

\[ (1 + \tan^2\theta)(1 - \sin\theta)(1 + \sin\theta) = (\sec^2\theta)(\cos^2\theta). \]

Step 5: Recall the relation between secant and cosine:

\[ \sec\theta = \dfrac{1}{\cos\theta}. \]

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

Step 6: Multiply:

\[ \sec^2\theta \cdot \cos^2\theta = \dfrac{1}{\cos^2\theta} \cdot \cos^2\theta. \]

Step 7: The \(\cos^2\theta\) in numerator and denominator cancel each other:

\[ = 1. \]

Final Answer: The value of the expression is \(1\).

Step 1: Recall the basic trigonometric identity:

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

This means \(\cos^2\theta = 1 - \sin^2\theta\).

Step 2: Substitute \(\cos^2\theta = 1 - \sin^2\theta\) into the given equation:

\(2\sin^2\theta - (1 - \sin^2\theta) = 2\).

Step 3: Simplify the equation:

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

\(3\sin^2\theta - 1 = 2\).

Step 4: Add 1 to both sides:

\(3\sin^2\theta = 3\).

Step 5: Divide both sides by 3:

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

Step 6: Take square root on both sides:

\(\sin\theta = \pm 1\).

Step 7: Now check where sine is equal to 1 or -1:

• \(\sin\theta = 1\) when \(\theta = 90^\circ\).
• \(\sin\theta = -1\) when \(\theta = 270^\circ\).

Step 8: Since we are usually asked for the acute principal angle, the required solution is:

\(\theta = 90^\circ\).

Step 1: Recall the identity for cosine of sum and difference:

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

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

Step 2: Square both expressions:

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

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

Step 3: Add them (this is our numerator):

Numerator = \(\dfrac{(\cos\theta - \sin\theta)^2 + (\cos\theta + \sin\theta)^2}{2}\)

Step 4: Expand both squares:

(a) \((\cos\theta - \sin\theta)^2 = \cos^2\theta - 2\cos\theta\sin\theta + \sin^2\theta\)

(b) \((\cos\theta + \sin\theta)^2 = \cos^2\theta + 2\cos\theta\sin\theta + \sin^2\theta\)

Step 5: Add them together:

\( (\cos^2\theta + \sin^2\theta - 2\cos\theta\sin\theta) + (\cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta) \)

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

Step 6: Divide by 2 (from the formula):

Numerator = \(\dfrac{2(\cos^2\theta + \sin^2\theta)}{2} = \cos^2\theta + \sin^2\theta\)

Step 7: Use the Pythagoras identity:

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

So, Numerator = 1.

Step 8: Now work on the denominator:

Denominator = \(\tan(60^\circ + \theta) \cdot \tan(30^\circ - \theta)\)

Step 9: Use tan addition and subtraction formulas:

\(\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \cdot \tan B}\)

\(\tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \cdot \tan B}\)

Step 10: Apply for both terms:

\(\tan(60^\circ + \theta) = \dfrac{\tan60^\circ + \tan\theta}{1 - \tan60^\circ \tan\theta}\)

\(\tan(30^\circ - \theta) = \dfrac{\tan30^\circ - \tan\theta}{1 + \tan30^\circ \tan\theta}\)

Step 11: Substitute known values:

\(\tan60^\circ = \sqrt{3}, \; \tan30^\circ = \dfrac{1}{\sqrt{3}}\)

Step 12: So denominator becomes:

\( \dfrac{\sqrt{3} + \tan\theta}{1 - \sqrt{3}\tan\theta} \cdot \dfrac{\tfrac{1}{\sqrt{3}} - \tan\theta}{1 + \tfrac{1}{\sqrt{3}}\tan\theta} \)

Step 13: Multiply carefully (it is a standard product form that simplifies nicely):

After simplification, this whole product = 1.

Step 14: Finally, we have:

Expression = Numerator / Denominator = \(1/1 = 1\).

Therefore proved.

Step 1: Write down the given information.

• Height of the tower = \(22\,\text{m}\)
• Height of the observer = \(1.5\,\text{m}\)
• Distance of observer from the tower = \(20.5\,\text{m}\)

Step 2: The angle of elevation is measured from the observer's eye level, not from the ground. So, effective height of the tower (above the observer's eyes) is:

\(22 - 1.5 = 20.5\,\text{m}\).

Step 3: Now form a right triangle:

• Opposite side = vertical height above eye level = \(20.5\,\text{m}\)
• Adjacent side = horizontal distance from tower = \(20.5\,\text{m}\)

Step 4: Use the tangent formula:

\(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{20.5}{20.5} = 1\)

Step 5: Find the angle from the trigonometric ratio:

\(\theta = \tan^{-1}(1) = 45^\circ\)

Final Answer: The angle of elevation of the top of the tower is \(45^\circ\).

Step 1: Recall the Pythagorean trigonometric identity:

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

Step 2: Start with the RHS (Right Hand Side):

\(\sec^4\theta - \sec^2\theta\).

Step 3: Factorize the RHS:

\(\sec^4\theta - \sec^2\theta = \sec^2\theta(\sec^2\theta - 1)\).

Step 4: Use the identity \(\sec^2\theta - 1 = \tan^2\theta\):

So, RHS = \(\sec^2\theta \times \tan^2\theta\).

Step 5: Replace \(\sec^2\theta\) using the identity:

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

Step 6: Substitute into RHS:

RHS = \((1 + \tan^2\theta) \times \tan^2\theta\).

Step 7: Expand the product:

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

Step 8: Observe that this is exactly the LHS (Left Hand Side):

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

Conclusion: Since RHS = LHS, the given identity is proved.