NCERT Solutions for Exercise 8.2 Class 10 Maths Chapter 8 - Introduction to Trigonometry

$(i) \sin 60^{o} \cos 30^{o} + \sin 30^{o} \cos 60^{o}$

Answer:

$\sin 60^{o} \cos 30^{o} + \sin 30^{o} \cos 60^{o}$
As we know,
the value of $\sin 60^{0} = \sqrt{3} / 2 = \cos 30^{0}$ , $\sin 30^{0} = 1 / 2 = \cos 60^{0}$
$\Rightarrow \frac{\sqrt{3}}{2} . \frac{\sqrt{3}}{2} + \frac{1}{2} . \frac{1}{2}$
$= \frac{3}{4} + \frac{1}{4}$
$= 1$

$(i i) 2 \tan^{2} 45^{o} + 2 \cos^{2} 30^{o} - 2 \sin^{2} 60^{o}$

Answer:

We know the value of

$\tan 45^{0} = 1$ and

$\cos 30^{0} = \sin 60^{0} = \frac{\sqrt{3}}{2}$
According to question,

$= 2 \tan^{2} 45^{o} + 2 \cos^{2} 30^{o} - 2 \sin^{2} 60^{o}$
$= 2 (1)^{2} + (\frac{\sqrt{3}}{2})^{2} - (\frac{\sqrt{3}}{2}) = 2$

$(i i i) \frac{\cos 45^{o}}{\sec 30^{o} + \csc 30^{o}}$

Answer:

$\frac{\cos 45^{o}}{\sec 30^{o} + \csc 30^{o}}$
we know the value of

$\cos 45 = 1 / \sqrt{2}$ , $\sec 30^{0} = 2 / \sqrt{3}$ and $c o s e c 30 = 2$ ,

After putting these values
$= \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2}$
$= \frac{1 / \sqrt{3}}{(2 + 2 \sqrt{3}) / \sqrt{3}}$
$= \frac{\sqrt{3}}{2 \sqrt{2} + 2 \sqrt{6}} \times \frac{2 \sqrt{2} - 2 \sqrt{6}}{2 \sqrt{2} - 2 \sqrt{6}}$
$= \frac{2 \sqrt{6} - 2 \sqrt{18}}{- 16} = 2 \frac{\sqrt{6} - 3 \sqrt{3}}{- 16} = \frac{3 \sqrt{3} - \sqrt{6}}{8}$

$(i v) \frac{\sin 30^{o} + \tan 45^{o} - c o s e c 60^{o}}{\sec 30^{o} + \cos 60^{o} + \cot 45^{o}}$

Answer:

$\frac{\sin 30^{o} + \tan 45^{o} - c o s e c 60^{o}}{\sec 30^{o} + \cos 60^{o} + \cot 45^{o}}$ ..................(i)
It is known that the values of the given trigonometric functions,
$\sin 30^{0} = 1 / 2 = c o s 60^{0} \tan 45^{0} = 1 = \cot 45^{0} \sec 30^{0} = 2 / \sqrt{3} = c o s e c 60^{0}$
Put all these values in equation (i), we get;
$\Rightarrow \frac{1 / 2 + 1 - 2 / \sqrt{3}}{2 / \sqrt{3} + 1 / 2 + 1} \Rightarrow \frac{3 / 2 - 2 / \sqrt{3}}{3 / 2 + 2 / \sqrt{3}} \Rightarrow \frac{3 \sqrt{3} - 4}{3 \sqrt{3} + 4} \times \frac{4 - 3 \sqrt{3}}{4 - 3 \sqrt{3}} \Rightarrow \frac{12 \sqrt{3} - 27 - 16 + 12 \sqrt{3}}{- 11} \Rightarrow \frac{43 - 24 \sqrt{3}}{11}$

Q 3: If $\tan (A + B) = \sqrt{3}$ and $\tan (A - B) = \frac{1}{\sqrt{3}};$ $0^{o} < A + B \leq 90^{o}; A > B,$ find $A a n d B .$

$(v) \frac{5 \cos^{2} 60^{o} + 4 \sec^{2} 30^{o} - \tan^{2} 45^{o}}{\sin^{2} 30^{o} + \cos^{2} 30^{o}}$

Answer:

$\frac{5 \cos^{2} 60^{o} + 4 \sec^{2} 30^{o} - \tan^{2} 45^{o}}{\sin^{2} 30^{o} + \cos^{2} 30^{o}}$ .....................(i)
We know the values of-
$\cos 60^{0} = 1 / 2 = \sin 30^{0} \sec 30^{0} = 2 / \sqrt{3} \tan 45^{0} = 1 \cos 30^{0} = \sqrt{3} / 2$
By substituting all these values in equation(i), we get;

$\Rightarrow \frac{5. (\frac{1}{2})^{2} + 4. (\frac{2}{\sqrt{3}})^{2} - 1}{(\frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} \Rightarrow \frac{5 / 4 - 1 + 16 / 3}{1} \Rightarrow \frac{1 / 4 + 16 / 3}{1} \Rightarrow \frac{67}{12}$

Q 2: Choose the correct option and justify your choice :

$(i) \frac{2 \tan 30^{o}}{1 + \tan^{2} 30^{o}} =$

$(A) \sin 60^{o}$ $(B) \cos 60^{o}$ $(C) \tan 60^{o}$ $(D) \sin 30^{o}$

Answer:

Put the value of tan 30 in the given question-
$\frac{2 \tan 30^{o}}{1 + \tan^{2} 30^{o}} = \frac{2 \times 1 / \sqrt{3}}{1 + (1 / \sqrt{3})^{2}} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \sqrt{3} / 2 = \sin 60^{0}$

The correct option is (A)

Q 2: Choose the correct option and justify your choice :

$(i i) \frac{1 - \tan^{2} 45^{o}}{1 + \tan^{2} 45^{o}} =$

$(A) \tan 90^{o}$ $(B) 1$ $(C) \sin 45^{o}$ $(D) 0$

Answer:

The correct option is (D)
$\frac{1 - \tan^{2} 45^{o}}{1 + \tan^{2} 45^{o}} =$
We know that $\tan 45 = 1$
So, $\frac{1 - 1}{1 + 1} = 0$

Q 2: Choose the correct option and justify your choice :

$(i i i) \sin 2 A = 2 \sin A$ is true when $A$ =

$(A) 0^{o}$ $(B) 30^{o}$ $(C) 45^{o}$ $(D) 60^{o}$

Answer:

The correct option is (A)
$\sin 2 A = 2 \sin A$
We know that $\sin 2 A = 2 \sin A \cos A$
So, $2 \sin A \cos A = 2 \sin A$
$\cos A = 1 A = 0^{0}$

Q 2: Choose the correct option and justify your choice :

$(i v) \frac{2 \tan 30^{o}}{1 - t a n^{2} 30^{o}} =$

$(A) \cos 60^{o}$ $(B) \sin 60^{o}$ $(C) \tan 60^{o}$ $(D) \sin 30^{o}$

Answer:

Put the value of $t a n 30^{o} = \1 / \sqrt{3}$

$= \frac{2 \tan 30^{o}}{1 - t a n^{2} 30^{o}} = \frac{2 \times 1 / \sqrt{3}}{1 - (\frac{1}{\sqrt{3}})^{2}} = \frac{2 / \sqrt{3}}{1 - 1 / 3} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} = \tan 60^{0}$

The correct option is (C)

Answer:

Given that,
$\tan (A + B) = \sqrt{3} = \tan 60^{0}$
So, $A + B = 60^{o}$ ..........(i)
$\tan (A - B) = 1 / \sqrt{3} = \tan 30^{0}$
therefore, $A - B = 30^{o}$ .......(ii)
By solving the equation (i) and (ii) we get;

$A = 45^{o}$ and $B = 15^{o}$

Q 4: State whether the following are true or false. Justify your answer.

$(i) \sin (A + B) = \sin A + \sin B$
$(i i)$ The value of $\sin θ$ increases as $θ$ increases.
$(i i i)$ The value of $\cos θ$ increases as $θ$ increases.
$(i v) \sin θ = \cos θ$ for all values of $θ$ .
$(v) \cot A$ is not defined for $A = 0^{o}$

Answer:

(i) False,
Let A = B = $45^{0}$
Then, $\sin (45^{0} + 45^{0}) = \sin 45^{0} + \sin 45^{0} \sin 90^{=} 1 / \sqrt{2} + q / \sqrt{2} 1 \neq \sqrt{2}$

(ii) True,
Take $θ = 0^{0}, 30^{0}, 45^{0}$
whent
$θ$ = 0 then zero(0),
$θ$ = 30 then value of $\sin θ$ is 1/2 = 0.5
$θ$ = 45 then value of $\sin θ$ is 0.707

(iii) False,
$\cos 0^{0} = 1, \cos 30^{0} = \sqrt{3} / 2 = 0.87, \cos 45^{0} = 1 \sqrt{2} = 0.707$

(iv) False,
Let $θ$ = 0
$\sin 0^{0} = \cos 0^{0} 0 \neq 1$

(v) True,
$\cot 0^{0} = \frac{\cos 0^{0}}{\sin 0^{0}} = \frac{1}{0}$ (not defined)

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Topics Covered in Chapter 8, Introduction to Trigonometry: Exercise 8.2

1. Trigonometric Ratios of Standard Angles: Evaluating sine, cosine, and tangent for 0°, 30°, 45°, 60°, and 90°.

2. Complementary Angles: Understanding the relationship between trigonometric ratios of complementary angles.

3. Application of Trigonometric Values: The practice of solving problems requires applying standard trigonometric values.

4. True or False Statements: The assessment deals with true or false statements about trigonometric ratios to confirm their validity while providing justification for the correct answers.

5. Multiple-Choice Questions: Students must choose appropriate trigonometric ratios from lists of options through calculations.

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Frequently Asked Questions (FAQs)

1. What is the value of cos60° sin30° + sin60°cos30° ?

cos60° sin30° + sin60°cos30°

=1/4+3/4=4/4

2. What is the value of cos 3 π?

The value of cos 3π is cos(π+2π)

=cos(π) =-1

3. What is the value of tan 45° ?

The value of tan 45° is 1

4. State true or false. The trigonometric function that is equal to the ratio of the side opposite a given angle to the hypotenuse is called sine.

Yes, the given assertion is correct. The sine is the ratio of the opposing side to the hypotenuse of a right-angle triangle.

Sin θ denotes the opposite side/hypotenuse

5. Say yes/no. Whether the value of cos 30° is equal to sin 60°

Yes, the value of cos 30° is equal to sin 60°.

6. What is the value of cosec 0° ?

The value of cosec 0° is not defined.

7. According to NCERT solutions for Class 10 Maths chapter 8 exercise 8.2 , What are the trigonometric ratios of specific angles?

The Trigonometric ratios of some specific angles are the ratios of the sides of a right-angle triangle with regard to any of its acute angles.

8. According to NCERT solutions for Class 10 Maths chapter 8 exercise 8.2 , what are the six trigonometric ratios ?

The six trigonometric ratios are Sine , Cosine , Tangent , Cotangent , Cosecant and Secant.

9. How many questions are there in the NCERT solutions for Class 10 Maths chapter 8 exercise 8.2 and what types of questions are there ?

NCERT solutions for Class 10 Maths chapter 8 exercise sides 8.2 consist of 4 questions and all the 4 questions are based on trigonometric ratios of specific angles.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

i studied 1 to 10th CBSE board abroad and 11th 12th in karnataka state board . i am a domicile of karnataka and comes under minority, income under 5 lakhs. how neet councillng will allocate my seat

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

show the previous year exams ?

Mitali Sharma 30 January,2025

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

SWETCHCHA MISKA 10 July,2024

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

Manasvini Gupta 23 July,2024

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

Ashvadha 03 July,2024

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.