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NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Edited By Ramraj Saini | Updated on Nov 27, 2023 09:23 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.4

NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 Introduction to Trigonometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 8.4 exercise is one of the most important exercises in trigonometry both for exam and for aptitude. When an equation holds true for all possible values of the variables, it is referred to as an identity. Similarly, a trigonometric identity is an equation involving trigonometric ratios of an angle that holds true for all values of the angle(s).

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  1. NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.4
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 8 exercise 8.4
  3. Assess NCERT Solutions for Class 10 Maths chapter 8 exercise 8.4
  4. More About NCERT Solutions for Class 10 Maths Exercise 8.4
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 8.4
  6. NCERT Solutions Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry
NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

NCERT solutions for exercise 8.4 Class 10 Maths chapter 8 Introduction to Trigonometry focuses on trigonometry identity. Also used Pythagoras theorem to prove these identuty and problems. 10th class Maths exercise 8.4 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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NCERT Solutions for Exercise 8.4

Q1 Express the trigonometric ratios \sin A,\sec A and \tan A in terms of cot A .

Answer:

We know that \csc^2A -\cot^2A = 1
(i)
\\\Rightarrow \frac{1}{\sin^2A}= 1+\cot^2A\\ \Rightarrow\sin^2A = \frac{1}{1+\cot^2A}\\ \Rightarrow \sin A = \frac{1}{\sqrt{1+\cot^2A}}

(ii) We know the identity of
1635934455463

(iii) \tan A = \frac{1}{\cot A}

Q2 Write all the other trigonometric ratios of \angle A in terms of \sec A .

Answer:

We know that the identity \sin^2 A + \cos^2 =1
\\\sin^2 A =1- \cos^2 \\ sin^2A = 1-\frac{1}{\sec^2A}
=\frac{\sec^2A -1}{\sec^2A}
\sin A =\sqrt{\frac{\sec^2A -1}{\sec^2A}}
=\frac{{}\sqrt{\sec^2A -1}}{\sec A}

cosec A =\frac{\sec A}{\sqrt{\sec^2A -1}}

\tan A = \frac{\sin A}{\cos A} = \sqrt{\sec^2A -1}

\cot A =\frac{1}{ \sqrt{\sec^2A -1}}

Q3 Evaluate :

(i)\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}

Answer:

\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}} ....................(i)

The above equation can be written as;

\\=\frac{\sin ^{2}63^{o}+\sin ^{2}(90^0-63^{o})}{\cos ^{2}(90^0-73^{o})+\cos ^{2}73^{o}}\\\\ =\frac{\sin ^{2}63^{o}+\cos ^{2}63^{o}}{\sin ^{2}73^{o}+\cos ^{2}73^{o}}\\\\ = 1
(Since \sin^2\theta +\cos^2\theta = 1 )

Q3 Evaluate :

(ii)\sin 25^{o}\cos 65^{o}+\cos 25^{o}\sin 65^{o}

Answer:

\sin 25^{o}\cos 65^{o}+\cos 25^{o}\sin 65^{o}

We know that
\\\sin(90^0-\theta) = \cos \theta \\\cos (90^0-\theta) = \sin \theta

Therefore,

\\\sin 25^{o}\cos (90^0-25^{o})+\cos 25^{o}\sin (90^0-25^{o})\\ \sin 25^0.\sin 25^0 + \cos 25^0.\cos 25^0\\ sin^2 25^0+\cos ^225^0\\ 1

Q4 Choose the correct option. Justify your choice.

(i) 9\sec^{2}A-9\tan^{2}A=

(A) 1 (B) 9 (C) 8 (D) 0

Answer:

The correct option is (B) = 9

9\sec^2A-9 \tan ^2A = 9(\sec^2A- \tan ^2A) .............(i)

and it is known that sec2A-tan2A=1

Therefore, equation (i) becomes, 9\times 1 = 9

Q4 Choose the correct option. Justify your choice.

(ii)(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\: \theta )=

(A) 0 (B) 1 (C) 2 (D) –1

Answer:

The correct option is (C)

(1+\tan \theta +\sec \theta )(1+\cot \theta -cosec\: \theta ) .......................(i)

we can write his above equation as;
\\=(1+\sin \theta/\cos \theta +1/\cos \theta )(1+\cos\theta/\sin \theta -1/sin\theta )\\\\= \frac{(1+\sin\theta+\cos\theta)}{\cos\theta.\sin\theta}\times (\frac{(\sin\theta+\cos\theta-1}{\sin\theta.\cos\theta})\\\\= \frac{(\sin\theta+\cos\theta)^2-1^2}{\sin\theta.\cos\theta}\\\\= \frac{\sin^2\theta+\cos^2\theta+2\sin\theta.\\cos\theta-1}{\sin\theta.\cos\theta}\\\\= 2\times\frac{\sin\theta.\cos\theta}{\sin\theta.\cos\theta}
= 2

Q4 Choose the correct option. Justify your choice.

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

(A)\sec A (B)\sin A (C)cosec A (D) \cos A

Answer:

The correct option is (D)

(\sec A+\tan A)(1-\sin A)=
\\ \Rightarrow ( \frac{1}{\cos A}+\frac{\sin A}{\cos A})(1-\sin A)\\\\ \Rightarrow \frac{1+\sin A}{\cos A}(1-\sin A)\\\\ \Rightarrow \frac{1-\sin^2 A}{\cos A}\\\\\Rightarrow \cos A

Q4 Choose the correct option. Justify your choice.

(iv) \frac{1+\tan ^{2}A}{1+\cot ^{2}A}=

(A) \sec ^{2}A (B) -1 (C) \cot ^{2}A (D) \tan ^{2}A

Answer:

The correct option is (D)

\frac{1+\tan ^{2}A}{1+\cot ^{2}A} ..........................eq (i)

The above equation can be written as;

We know that \cot A = \frac{1}{\tan A}

therefore,

\\\Rightarrow \frac{1+\tan ^{2}A}{1+\frac{1}{\tan ^{2}}A}\\ \Rightarrow \tan^2A\times(\frac{1+\tan^2 A}{1+\tan^2A})\\ \Rightarrow \tan^2A

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }

Answer:

We need to prove-
(\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }

Now, taking LHS,

(\csc\theta-\cot\theta)^2 = (\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta})^2
\\= (\frac{1-\cos\theta}{\sin\theta})^2\\ =\frac{(1-\cos\theta)(1-\cos\theta)}{\sin^2\theta}\\
\\=\frac{(1-\cos\theta)(1-\cos\theta)}{1-\cos^2\theta}\\\\ =\frac{(1-\cos\theta)(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\\\ =\frac{1-\cos\theta}{1+\cos\theta}

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(ii) \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A

Answer:

We need to prove-

\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}= 2\sec A

taking LHS;
\\=\frac{\cos^2A+1+\sin^2A+2\sin A}{\cos A(1+\sin A)}\\\\ =\frac{2(1+\sin A)}{\cos A(1+\sin A)}\\\\ =2/\cos A = 2\sec A

= RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(iii)\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta

[ Hint: Write the expression in terms of \sin \theta and \cos\theta ]

Answer:

We need to prove-
\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \:cosec \theta

Taking LHS;

\\\Rightarrow \frac{\tan^2 \theta }{\tan \theta-1 }+\frac{1}{\tan\theta(1-\tan \theta) }\\\\\ \Rightarrow\frac{\tan^3\theta-\tan^4\theta+\tan\theta-1}{(\tan\theta-1).\tan\theta.(1-\tan\theta)}\\\\ \Rightarrow \frac{(\tan^3\theta-1)(1-\tan\theta)}{\tan\theta.(\tan\theta-1)(1-\tan\theta)}\\
By using the identity a 3 - b 3 =(a - b) (a 2 + b 2 +ab)

\\\Rightarrow \frac{(\tan\theta -1)(\tan^2\theta+1+\tan\theta)}{\tan\theta(\tan\theta -1a)}\\\\ \Rightarrow \tan\theta+1+\frac{1}{\tan\theta}\\\\ \Rightarrow 1+\frac{1+\tan^2\theta}{\tan\theta}\\\\ \Rightarrow 1+\sec^2\theta \times \frac{1}{\tan\theta}\\\\ \Rightarrow 1+\sec\theta.\csc\theta\\\\ =RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(iv)\frac{1+\sec A}{\sec A}=\frac{\sin ^{2}A}{1-\cos A}

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-
\frac{1+\sec A}{\sec A}=\frac{\sin ^{2}A}{1-\cos A}

taking LHS;

\\\Rightarrow \frac{1+\sec A}{\sec A}\\ \Rightarrow (1+\frac{1}{\cos A})/\sec A\\ \Rightarrow 1+\cos A

Taking RHS;
We know that identity 1-\cos^2\theta = \sin^2\theta

\\\Rightarrow \frac{\sin ^{2}A}{1-\cos A}\\ \Rightarrow \frac{1-\cos^2 A}{1-\cos A}\\ \Rightarrow \frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}\\ \Rightarrow 1+\cos A

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (v) \frac{\cos A-\sin A+1}{\cos A+\sin A-1}= \csc A+\cot A , using the identity \csc ^{2}A= 1+\cot ^{2}A

Answer:

We need to prove -
\frac{\cos A-\sin A+1}{\cos A+\sin A-1}= cosec A+\cot A

Dividing the numerator and denominator by \sin A , we get;

\\=\frac{\cot A-1+\csc A}{\cot A +1-\csc A}\\\\= \frac{(\cot A+\csc A)-(\csc^2 A-\cot^2A)}{\cot A +1-\csc A}\\\\= \frac{(\csc A+\cot A)(1-\csc A+\cot A)}{\cot A +1-\csc A}\\\\= \csc A+\cot A\\\\ =RHS

Hence Proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(vi)\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A

Answer:

We need to prove -
\sqrt{\frac{1+\sin A}{1-\sin A}}= \sec A+\tan A
Taking LHS;
By rationalising the denominator, we get;

\\= \sqrt{\frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}\\\\ = \sqrt{\frac{(1+\sin A)^2}{1-\sin^2A}}\\\\ =\frac{1+\sin A}{\cos A}\\\\ = \sec A + \tan A\\\\ = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii)\frac{\sin \theta -2\sin ^{3}\theta }{2\cos ^{3}\theta -\cos \theta }= \tan \theta

Answer:

We need to prove -
\frac{\sin \theta -2\sin ^{3}\theta }{2\cos ^{3}\theta -\cos \theta }= \tan \theta

Taking LHS;
[we know the identity \cos2\theta = 2\cos^2\theta-1=\cos^2\theta-\sin^2\theta ]

\\\Rightarrow \frac{\sin \theta(1 -2\sin ^{2}\theta) }{\cos\theta(2\cos ^{2}\theta -1) }\\\\ \Rightarrow \frac{\sin\theta(\sin^2\theta+\cos^2\theta-2\sin^\theta)}{\cos\theta.\cos2\theta}\\\\ \Rightarrow \frac{\sin\theta.\cos2\theta}{\cos\theta.\cos2\theta}\\\\ \Rightarrow \tan\theta =RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(viii)(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A

Answer:

Given equation,
(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}= 7+\tan ^{2}A+\cot ^{2}A ..................(i)

Taking LHS;

(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}
\\\Rightarrow \sin^2 A+\csc^2A +2+\cos^2A+\sec^2A+2\\\\ \Rightarrow 1+2+2+(1+\cot^2A)+(1+\tan^2A)
[since \sin^2\theta +\cos^2\theta = 1, \csc^2\theta-\cot^2\theta =1, \sec^2\theta-\tan^2\theta=1 ]

\\7+\csc^2A+\tan^2A\\ =RHS

Hence proved

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(ix)\:(cosec A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-
(coesc A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}
Taking LHS;
\\\Rightarrow (cosec A-\frac{1}{\csc A})(\sec A-\frac{1}{\sec A})\\\\ \Rightarrow\frac{(cosec^2-1)}{cosec A}\times\frac{\sec^2A-1}{\sec A}\\\\ \Rightarrow\frac{\cot^2A}{cosec A}.\frac{\tan^2A}{\sec A}\\\\ \Rightarrow\sin A .\cos A

Taking RHS;

\\\Rightarrow\frac{1}{\sin A/\cos A+\cos A/\sin A}\\\\ \Rightarrow\frac{\sin A .\cos A}{\sin^2A+\cos^2A}\\\\ \Rightarrow \sin A.\cos A

LHS = RHS

Hence proved.

Q5 Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(x) (\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A

Answer:

We need to prove,
(\frac{1+\tan ^{2}A}{1+\cot ^{2}A})= (\frac{1-\tan A}{1-\cot A})^{2}= \tan ^{2}A

Taking LHS;

\\\Rightarrow \frac{1+\tan ^{2}A}{1+\cot ^{2}A} = \frac{\sec^2A}{\csc^2A}=\tan^2A

Taking RHS;

\\=(\frac{1-\tan A}{1-\cot A})^2\\\\= (\frac{1-\sin A/\cos A }{1-\cos A /\sin A})^2\\\\ = \frac{(\cos A -\sin A)^2(\sin^2A)}{(\sin A-\cos A)^2(\cos^2A)}\\\\ =\tan^2A

LHS = RHS

Hence proved.

More About NCERT Solutions for Class 10 Maths Exercise 8.4

NCERT solutions for Class 10 Maths exercise 8.4- We will try to derive an important formula from NCERT solutions for Class 10 Maths chapter 8 exercise 8.4 by applying the Pythagoras theorem to it. And the Pythagoras Theorem states that the sum of the squares of the base and the perpendicular is equal to the square of the hypotenuse.

.

We will divide the whole equation by

And we know that sin is the ratio of perpendicular and hypotenuse while cos is the ratio of base and hypotenuse.for all values of angle, A lying between 0° and 90°.

The other formulas that have similar use as the above are:and Students can use these Introduction to Trigonometry Class 10 notes to quick revision of the important concepts discussed in this chapter.

Benefits of NCERT Solutions for Class 10 Maths Exercise 8.4

  • Exercise 8.4 Class 10 Maths, is based on the main concept of Trigonometric Identities.
  • NCERT syllabus Class 10 Maths chapter 8 exercise 8.4 helps in solving and revising all questions of these exercises.

Mastering the values of these trigonometric identities of Class 10 Maths chapter 8 exercise 8.4 can help in simplifying complex trigonometric questions into simpler ones and proving different equations of physics in later higher classes.

Also see-

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