Class 10 Maths Chapter 8 Exercise 8.2 NCERT Solutions

Question 1 NCERT Ex 8.2

Evaluate the following expressions:

(i) $sin 30^circ cos 60^circ + cos 30^circ sin 60^circ$
(ii) $3 an^2 45^circ - sin^2 60^circ - rac{1}{2}cot^2 30^circ$
(iii) $rac{cos 60^circ}{sec 45^circ + ext{cosec } 45^circ}$
(iv) $rac{4 sin^2 60^circ + 3 an^2 30^circ - 8 sin 45^circ cos 45^circ}{sin^2 30^circ + cos^2 30^circ}$

Part (i): $sin 30^circ cos 60^circ + cos 30^circ sin 60^circ$

Substitute standard values: $sin 30^circ = rac{1}{2}$, $cos 60^circ = rac{1}{2}$, $cos 30^circ = rac{sqrt{3}}{2}$, $sin 60^circ = rac{sqrt{3}}{2}$. $$= left(rac{1}{2} ight)left(rac{1}{2} ight) + left(rac{sqrt{3}}{2} ight)left(rac{sqrt{3}}{2} ight) = rac{1}{4} + rac{3}{4} = rac{4}{4} = 1$$

Part (ii): $3 an^2 45^circ - sin^2 60^circ - rac{1}{2}cot^2 30^circ$

Substitute values: $ an 45^circ = 1$, $sin 60^circ = rac{sqrt{3}}{2}$, $cot 30^circ = sqrt{3}$. $$= 3(1)^2 - left(rac{sqrt{3}}{2} ight)^2 - rac{1}{2}(sqrt{3})^2 = 3(1) - rac{3}{4} - rac{3}{2}$$ $$= 3 - rac{3}{4} - rac{6}{4} = 3 - rac{9}{4} = rac{12 - 9}{4} = rac{3}{4}$$

Part (iii): $rac{cos 60^circ}{sec 45^circ + ext{cosec } 45^circ}$

Substitute: $cos 60^circ = rac{1}{2}$, $sec 45^circ = sqrt{2}$, $ ext{cosec } 45^circ = sqrt{2}$. $$= rac{rac{1}{2}}{sqrt{2} + sqrt{2}} = rac{rac{1}{2}}{2sqrt{2}} = rac{1}{4sqrt{2}}$$ Rationalizing the denominator: $$= rac{1 imes sqrt{2}}{4sqrt{2} imes sqrt{2}} = rac{sqrt{2}}{4 imes 2} = rac{sqrt{2}}{8}$$

Part (iv): $rac{4 sin^2 60^circ + 3 an^2 30^circ - 8 sin 45^circ cos 45^circ}{sin^2 30^circ + cos^2 30^circ}$

Denominator: $sin^2 30^circ + cos^2 30^circ = left(rac{1}{2} ight)^2 + left(rac{sqrt{3}}{2} ight)^2 = rac{1}{4} + rac{3}{4} = 1$.
Numerator: $$= 4left(rac{sqrt{3}}{2} ight)^2 + 3left(rac{1}{sqrt{3}} ight)^2 - 8left(rac{1}{sqrt{2}} ight)left(rac{1}{sqrt{2}} ight)$$ $$= 4left(rac{3}{4} ight) + 3left(rac{1}{3} ight) - 8left(rac{1}{2} ight) = 3 + 1 - 4 = 0$$ Therefore: $$rac{0}{1} = 0$$

Answers: (i) $1$, (ii) $rac{3}{4}$, (iii) $rac{sqrt{2}}{8}$, (iv) $0$.

Question 2 NCERT Ex 8.2

Choose the correct option and justify your choice:

(i) $rac{2 an 60^circ}{1 + an^2 60^circ} =$
(A) $sin 60^circ$    (B) $cos 60^circ$    (C) $ an 60^circ$    (D) $sin 30^circ$

(ii) $rac{1 - an^2 60^circ}{1 + an^2 60^circ} =$
(A) $ an 60^circ$    (B) $1$    (C) $sin 45^circ$    (D) $-rac{1}{2}$

(iii) $sin 2A = 2 sin A$ is true when $A =$
(A) $0^circ$    (B) $30^circ$    (C) $45^circ$    (D) $60^circ$

(iv) $rac{2 an 60^circ}{1 - an^2 60^circ} =$
(A) $cos 60^circ$    (B) $sin 60^circ$    (C) $ an 60^circ$    (D) $-sqrt{3}$

Part (i)

$$rac{2 an 60^circ}{1 + an^2 60^circ} = rac{2(sqrt{3})}{1 + (sqrt{3})^2} = rac{2sqrt{3}}{1 + 3} = rac{2sqrt{3}}{4} = rac{sqrt{3}}{2} = sin 60^circ$$ Hence, Option (A) is correct.

Part (ii)

$$rac{1 - an^2 60^circ}{1 + an^2 60^circ} = rac{1 - (sqrt{3})^2}{1 + (sqrt{3})^2} = rac{1 - 3}{1 + 3} = rac{-2}{4} = -rac{1}{2}$$ Hence, Option (D) is correct.

Part (iii)

Testing $A = 0^circ$:
$ ext{LHS} = sin(2 imes 0^circ) = sin 0^circ = 0$
$ ext{RHS} = 2 sin 0^circ = 2(0) = 0$
$ ext{LHS} = ext{RHS}$ when $A = 0^circ$.
For any other option (e.g., $A = 30^circ$), $sin 60^circ = rac{sqrt{3}}{2} eq 2left(rac{1}{2} ight) = 1$.
Hence, Option (A) is correct.

Part (iv)

$$rac{2 an 60^circ}{1 - an^2 60^circ} = rac{2(sqrt{3})}{1 - (sqrt{3})^2} = rac{2sqrt{3}}{1 - 3} = rac{2sqrt{3}}{-2} = -sqrt{3}$$ Hence, Option (D) is correct.

Answers: (i) (A) $sin 60^circ$, (ii) (D) $-rac{1}{2}$, (iii) (A) $0^circ$, (iv) (D) $-sqrt{3}$.

Question 3 NCERT Ex 8.2

If $sin(A + B) = 1$ and $cos(A - B) = rac{sqrt{3}}{2}$; $0^circ < A + B leq 90^circ$; $A > B$, find $A$ and $B$.

Step 1: Form Linear Equations in $A$ and $B$

From the first given condition: $$sin(A + B) = 1 implies A + B = 90^circ quad ext{...(i)}$$ From the second given condition: $$cos(A - B) = rac{sqrt{3}}{2} implies A - B = 30^circ quad ext{...(ii)}$$

Step 2: Solve the System of Equations

Adding equations (i) and (ii): $$(A + B) + (A - B) = 90^circ + 30^circ$$ $$2A = 120^circ implies A = rac{120^circ}{2} = 60^circ$$ Substituting $A = 60^circ$ into equation (i): $$60^circ + B = 90^circ implies B = 90^circ - 60^circ = 30^circ$$

Step 3: Check Conditions

$A + B = 60^circ + 30^circ = 90^circ$ (satisfies $0^circ < A + B leq 90^circ$) and $A = 60^circ > B = 30^circ$.

$A = 60^circ$ and $B = 30^circ$.

Question 4 NCERT Ex 8.2

State whether the following are true or false. Justify your answer:

(i) $ an(A + B) = an A + an B$.
(ii) The value of $cos heta$ increases as $ heta$ increases.
(iii) $sin heta = cos heta$ for all values of $ heta$.
(iv) $cot A$ is not defined for $A = 0^circ$.
(v) $sec heta$ is always greater than or equal to 1.

Part (i)

False. Take $A = 30^circ$ and $B = 30^circ$:
$ ext{LHS} = an(30^circ + 30^circ) = an 60^circ = sqrt{3}$.
$ ext{RHS} = an 30^circ + an 30^circ = rac{1}{sqrt{3}} + rac{1}{sqrt{3}} = rac{2}{sqrt{3}}$.
Since $sqrt{3} eq rac{2}{sqrt{3}}$, the equality does not hold in general.

Part (ii)

False. For standard angles from $0^circ$ to $90^circ$:
$cos 0^circ = 1$, $cos 30^circ = rac{sqrt{3}}{2} approx 0.866$, $cos 45^circ = rac{1}{sqrt{2}} approx 0.707$, $cos 60^circ = 0.5$, $cos 90^circ = 0$.
The value of $cos heta$ decreases as $ heta$ increases from $0^circ$ to $90^circ$.

Part (iii)

False. This is only true when $ heta = 45^circ$ ($sin 45^circ = cos 45^circ = rac{1}{sqrt{2}}$). For all other angles, $sin heta eq cos heta$ (e.g., $sin 30^circ = rac{1}{2}$ while $cos 30^circ = rac{sqrt{3}}{2}$).

Part (iv)

True. $cot 0^circ = rac{cos 0^circ}{sin 0^circ} = rac{1}{0}$, which is undefined (not defined).

Part (v)

True. For acute angles $ heta$ ($0^circ leq heta < 90^circ$), $sec heta = rac{ ext{Hypotenuse}}{ ext{Adjacent}}$. In any right triangle, the hypotenuse is the longest side, so $ ext{Hypotenuse} geq ext{Adjacent} implies sec heta geq 1$.

(i) False, (ii) False, (iii) False, (iv) True, (v) True.

Introduction to Trigonometry - Exercise 8.2

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