CBSE Class 10 Mathematics Trigonometry Assignment Set 08

Question 1. By taking any triangle of your choice, prove the following results.
(i) \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
(ii) \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)
(iii) \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \)
Answer:
(i) Let us consider an equilateral triangle \( ABC \) with each side of length \( 2a \). Each interior angle of this triangle measures \( 60^{\circ} \). We now draw a perpendicular line \( AD \) from vertex \( A \) to the base \( BC \). In this equilateral triangle, \( AD \) also bisects the angle \( A \) and the side \( BC \), giving us \( BD = a \) and \( \angle BAD = 30^{\circ} \).
By applying the Pythagorean theorem in the right-angled triangle \( ABD \):
\( AD^2 = AB^2 - BD^2 \)
\( AD^2 = (2a)^2 - a^2 = 4a^2 - a^2 = 3a^2 \)
\( AD = a\sqrt{3} \)
Now, using the definition of sine for \( \angle B = 60^{\circ} \) in the right-angled triangle \( ABD \):
\( \sin 60^{\circ} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AD}{AB} = \frac{a\sqrt{3}}{2a} = \frac{\sqrt{3}}{2} \).

(ii) Let us take an isosceles right-angled triangle \( ABC \) where \( \angle B = 90^{\circ} \) and the two equal sides are \( AB = BC = a \). Since the sides are equal, the opposite angles must be equal, so \( \angle A = \angle C = 45^{\circ} \).
Using the Pythagorean theorem, the hypotenuse is computed as follows:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = a^2 + a^2 = 2a^2 \)
\( AC = a\sqrt{2} \)
Now, using the cosine ratio for \( \angle C = 45^{\circ} \):
\( \cos 45^{\circ} = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} \).

(iii) Returning to the same equilateral triangle \( ABC \) of side \( 2a \) and altitude \( AD = a\sqrt{3} \) used in part (i), we focus on \( \angle BAD = 30^{\circ} \) in the right-angled triangle \( ABD \).
Using the tangent definition for \( 30^{\circ} \):
\( \tan 30^{\circ} = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BD}{AD} = \frac{a}{a\sqrt{3}} = \frac{1}{\sqrt{3}} \).
In simple words: To find these trigonometric values geometrically, we can construct simple triangles - an equilateral triangle split in half for \( 30^{\circ} \) and \( 60^{\circ} \) angles, and an isosceles right triangle for \( 45^{\circ} \) angles. This lets us use basic side-length ratios to prove the standard values.

Exam Tip: When proving standard trigonometric angles geometrically, always draw a neat diagram showing the side lengths \( a \) or \( 2a \) clearly. Don't forget to state the geometric properties you are using - such as why the altitude of an equilateral triangle bisects its base.

Evaluate the following (Q.No. 2 to Q.No. 6):

Question 2. \( 4(\sin^2 30^{\circ} + \cos^2 30^{\circ}) - 3\tan^2 45^{\circ} + 2\sec^2 60^{\circ} \)
Answer: We substitute the standard values of the trigonometric ratios:
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
\( \tan 45^{\circ} = 1 \)
\( \sec 60^{\circ} = 2 \)
Plugging these values into the given expression:
\( 4 \left[ \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \right] - 3(1)^2 + 2(2)^2 \)
\( = 4 \left( \frac{1}{4} + \frac{3}{4} \right) - 3(1) + 2(4) \)
\( = 4(1) - 3 + 8 \)
\( = 4 - 3 + 8 \)
\( = 9 \).
In simple words: Just swap each trigonometric term with its known fraction or number, square them if needed, and follow the standard order of operations to calculate the final value.

Exam Tip: Recall that \( \sin^2 \theta + \cos^2 \theta = 1 \) for any angle \( \theta \). Using this identity directly makes the first bracket \( 4(1) = 4 \) instantly and saves you time during exams.

Question 3. \( 3\cos^2 60^{\circ}\sec^2 30^{\circ} - 2\sin^2 30^{\circ}\tan^2 60^{\circ} \)
Answer: We substitute the standard trigonometric values:
\( \cos 60^{\circ} = \frac{1}{2} \)
\( \sec 30^{\circ} = \frac{2}{\sqrt{3}} \)
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \tan 60^{\circ} = \sqrt{3} \)
Let us put these values into the expression:
\( 3 \left(\frac{1}{2}\right)^2 \left(\frac{2}{\sqrt{3}}\right)^2 - 2 \left(\frac{1}{2}\right)^2 (\sqrt{3})^2 \)
\( = 3 \left(\frac{1}{4}\right) \left(\frac{4}{3}\right) - 2 \left(\frac{1}{4}\right) (3) \)
\( = 1 - \frac{6}{4} \)
\( = 1 - \frac{3}{2} \)
\( = -\frac{1}{2} \).
In simple words: Replace the functions with their standard values, compute the squares, and simplify the fractions to arrive at the result.

Exam Tip: Be very careful with signs and fractions when subtracting. Keep the denominators clean to avoid common arithmetic slip-ups under exam pressure.

Question 4. \( 4\tan^2 30^{\circ} - 2\cos^2 90^{\circ} + \frac{1}{2}\sin^2 90^{\circ} + \sin^2 30^{\circ}\cos^2 45^{\circ} \)
Answer: We use the following known ratios:
\( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \)
\( \cos 90^{\circ} = 0 \)
\( \sin 90^{\circ} = 1 \)
\( \sin 30^{\circ} = \frac{1}{2} \)
\( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)
Now we insert these values into our expression:
\( 4 \left( \frac{1}{\sqrt{3}} \right)^2 - 2(0)^2 + \frac{1}{2}(1)^2 + \left( \frac{1}{2} \right)^2 \left( \frac{1}{\sqrt{2}} \right)^2 \)
\( = 4 \left( \frac{1}{3} \right) - 0 + \frac{1}{2}(1) + \left( \frac{1}{4} \right) \left( \frac{1}{2} \right) \)
\( = \frac{4}{3} + \frac{1}{2} + \frac{1}{8} \)
To sum these terms, find the lowest common multiple (LCM) of 3, 2, and 8, which is 24:
\( = \frac{4 \times 8}{24} + \frac{1 \times 12}{24} + \frac{1 \times 3}{24} \)
\( = \frac{32 + 12 + 3}{24} \)
\( = \frac{47}{24} \).
In simple words: Insert the values of the angles, square each one, find a common denominator to add the fractions, and simplify the total.

Exam Tip: Showing the step where you calculate the LCM (in this case, 24) is crucial. Examiners like to see how you combine unlike fractions step-by-step.

Question 5. \( (\sin^2 90^{\circ} + \cos 45^{\circ} + \cos 60^{\circ}) \cdot (\cos^2 0^{\circ} - \sin 45^{\circ} + \sin 30^{\circ}) \)
Answer: We substitute the values of the trigonometric ratios:
\( \sin 90^{\circ} = 1 \implies \sin^2 90^{\circ} = 1 \)
\( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \cos 60^{\circ} = \frac{1}{2} \)
\( \cos 0^{\circ} = 1 \implies \cos^2 0^{\circ} = 1 \)
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \sin 30^{\circ} = \frac{1}{2} \)

Now, let us simplify each bracket:
First bracket:
\( 1 + \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{3}{2} + \frac{1}{\sqrt{2}} \)

Second bracket:
\( 1 - \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{3}{2} - \frac{1}{\sqrt{2}} \)

Multiplying the two brackets gives us an algebraic pattern of the form \( (a + b)(a - b) = a^2 - b^2 \), where \( a = \frac{3}{2} \) and \( b = \frac{1}{\sqrt{2}} \):
\( \left( \frac{3}{2} + \frac{1}{\sqrt{2}} \right) \left( \frac{3}{2} - \frac{1}{\sqrt{2}} \right) = \left( \frac{3}{2} \right)^2 - \left( \frac{1}{\sqrt{2}} \right)^2 \)
\( = \frac{9}{4} - \frac{1}{2} \)
\( = \frac{9}{4} - \frac{2}{4} \)
\( = \frac{7}{4} \).
In simple words: Evaluate each set of parentheses separately. You will end up with two expressions that look like \( A + B \) and \( A - B \). Multiply them easily by squaring the terms and subtracting.

Exam Tip: Look out for algebraic identities like \( (a+b)(a-b) = a^2-b^2 \) when simplifying product expressions. It is much faster and less prone to errors than multiplying each term out individually.

Question 6. \( \frac{\tan^2 60^{\circ} + 4\sin^2 45^{\circ} + 3\sec^2 30^{\circ} + 5\cos 90^{\circ}}{\text{cosec } 30^{\circ} + \sec 60^{\circ} - \cot^2 30^{\circ}} \)
Answer: Let us first write down and evaluate the numerator and denominator separately using standard values:

Numerator values:
\( \tan 60^{\circ} = \sqrt{3} \implies \tan^2 60^{\circ} = 3 \)
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \implies \sin^2 45^{\circ} = \frac{1}{2} \)
\( \sec 30^{\circ} = \frac{2}{\sqrt{3}} \implies \sec^2 30^{\circ} = \frac{4}{3} \)
\( \cos 90^{\circ} = 0 \)

Substituting these into the Numerator:
\( \text{Numerator} = 3 + 4\left(\frac{1}{2}\right) + 3\left(\frac{4}{3}\right) + 5(0) \)
\( = 3 + 2 + 4 + 0 = 9 \)

Denominator values:
\( \text{cosec } 30^{\circ} = 2 \)
\( \sec 60^{\circ} = 2 \)
\( \cot 30^{\circ} = \sqrt{3} \implies \cot^2 30^{\circ} = 3 \)

Substituting these into the Denominator:
\( \text{Denominator} = 2 + 2 - 3 = 1 \)

Now, dividing the numerator by the denominator:
\( \frac{\text{Numerator}}{\text{Denominator}} = \frac{9}{1} = 9 \).
In simple words: Work out the top part and the bottom part of the fraction separately by replacing each function with its standard value, then divide the top result by the bottom result.

Exam Tip: For complex rational expressions like this, separating the calculation into a distinct 'Numerator' and 'Denominator' section makes your working exceptionally neat and easy for the examiner to follow.

Question 7. Find \( \sin 15^{\circ} \), \( \cos 15^{\circ} \) from the relation \( \sin (A - B) = \sin A \cos B - \cos A \sin B \) and \( \cos (A - B) = \cos A \cos B + \sin A \sin B \)
Answer: Let us choose \( A = 45^{\circ} \) and \( B = 30^{\circ} \). Thus, \( A - B = 45^{\circ} - 30^{\circ} = 15^{\circ} \).
We know the values of the standard angles:
\( \sin 45^{\circ} = \frac{1}{\sqrt{2}} \), \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)
\( \sin 30^{\circ} = \frac{1}{2} \), \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)

1. To find \( \sin 15^{\circ} \), use the first relation:
\( \sin 15^{\circ} = \sin (45^{\circ} - 30^{\circ}) \)

\( \implies \sin 15^{\circ} = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \)
\( = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \)
\( = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} \)
\( = \frac{\sqrt{3} - 1}{2\sqrt{2}} \).

2. To find \( \cos 15^{\circ} \), use the second relation:
\( \cos 15^{\circ} = \cos (45^{\circ} - 30^{\circ}) \)

\( \implies \cos 15^{\circ} = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} \)
\( = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \)
\( = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} \)
\( = \frac{\sqrt{3} + 1}{2\sqrt{2}} \).
In simple words: We can rewrite \( 15^{\circ} \) as \( 45^{\circ} - 30^{\circ} \). Substituting these familiar angles into the formulas allows us to compute the values of \( \sin 15^{\circ} \) and \( \cos 15^{\circ} \) using their standard values.

Exam Tip: Be very careful with the plus and minus signs in compound angle formulas. Remember that \( \sin(A-B) \) has a minus sign, whereas \( \cos(A-B) \) has a plus sign.

Question 8. If \( \cos (A + B) = 0 \) and \( \sin (A - B) = \frac{\sqrt{3}}{2} \), find the angle A and B.
Answer: We are given the following two trigonometric equations:
1) \( \cos (A + B) = 0 \)
Since the cosine of \( 90^{\circ} \) is \( 0 \), we can set:
\( A + B = 90^{\circ} \) - Equation 1

2) \( \sin (A - B) = \frac{\sqrt{3}}{2} \)
Since the sine of \( 60^{\circ} \) is \( \frac{\sqrt{3}}{2} \), we can set:
\( A - B = 60^{\circ} \) - Equation 2

Now, let us solve these simultaneous linear equations:
Add Equation 1 and Equation 2:
\( (A + B) + (A - B) = 90^{\circ} + 60^{\circ} \)

\( \implies 2A = 150^{\circ} \)

\( \implies A = 75^{\circ} \)

Substitute the value of \( A = 75^{\circ} \) into Equation 1:
\( 75^{\circ} + B = 90^{\circ} \)

\( \implies B = 90^{\circ} - 75^{\circ} \)

\( \implies B = 15^{\circ} \).
Therefore, the angles are \( A = 75^{\circ} \) and \( B = 15^{\circ} \).
In simple words: Find which standard angles give a cosine of \( 0 \) and a sine of \( \frac{\sqrt{3}}{2} \). This gives you two simple equations for \( A \) and \( B \) that you can add together to find each angle's value.

Exam Tip: Always state clearly that the angles \( A \) and \( B \) are acute when working out standard class 10 problems, and verify your final values of \( A \) and \( B \) by plugging them back into the original equations.

Question 9. Find \( \tan 75^{\circ} \) from the relation \( \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
Answer: Let us set \( A = 45^{\circ} \) and \( B = 30^{\circ} \), so that \( A + B = 45^{\circ} + 30^{\circ} = 75^{\circ} \).
We know the standard tangent values:
\( \tan 45^{\circ} = 1 \)
\( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \)

Substituting these into the given formula:
\( \tan 75^{\circ} = \frac{\tan 45^{\circ} + \tan 30^{\circ}}{1 - \tan 45^{\circ} \tan 30^{\circ}} \)

\( \implies \tan 75^{\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - (1)\left(\frac{1}{\sqrt{3}}\right)} \)
\( = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} \)
\( = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)

Now, let us rationalize the denominator by multiplying both the numerator and the denominator by \( (\sqrt{3} + 1) \):
\( \tan 75^{\circ} = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \)
\( = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2} \)
\( = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} \)
\( = \frac{4 + 2\sqrt{3}}{2} \)
\( = 2 + \sqrt{3} \).
In simple words: Break \( 75^{\circ} \) down into \( 45^{\circ} + 30^{\circ} \) and use the given formula. After substituting the standard values, multiply the top and bottom of the fraction by \( \sqrt{3} + 1 \) to get rid of the roots in the denominator.

Exam Tip: Rationalizing the denominator is a mandatory step in trigonometry exams to get full marks. Leaving the answer as \( \frac{\sqrt{3}+1}{\sqrt{3}-1} \) will likely cost you a mark.

Question 10. Using the formula \( \sin 2A = \frac{2\tan A}{1 + \tan^2 A} \), find the value of \( \sin 60^{\circ} \) given that \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \).
Answer: We want to find the value of \( \sin 60^{\circ} \). Let us choose \( A = 30^{\circ} \), which gives \( 2A = 2(30^{\circ}) = 60^{\circ} \).
We are given that:
\( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \)

Now, substitute \( A = 30^{\circ} \) into the formula:
\( \sin 60^{\circ} = \frac{2\tan 30^{\circ}}{1 + \tan^2 30^{\circ}} \)

\( \implies \sin 60^{\circ} = \frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} \)
\( = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} \)
\( = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} \)
\( = \frac{2}{\sqrt{3}} \times \frac{3}{4} \)
\( = \frac{3}{2\sqrt{3}} \)
Rationalize the expression by multiplying numerator and denominator by \( \sqrt{3} \):
\( = \frac{3\sqrt{3}}{2 \times 3} \)
\( = \frac{\sqrt{3}}{2} \).
In simple words: Since we need the sine of \( 60^{\circ} \), we can set our angle \( A \) as \( 30^{\circ} \) in the given formula. Put in the value of \( \tan 30^{\circ} \), simplify the fraction, and you will get the final answer.

Exam Tip: Always double check that the final simplified value of any sine or cosine function is within the range of \( [-1, 1] \). Since \( \frac{\sqrt{3}}{2} \approx 0.866 \), this is a valid value and confirms your calculation is realistic.