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).

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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## Assess NCERT Solutions for Class 10 Maths chapter 8 exercise 8.4

NCERT Solutions for Exercise 8.4

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

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

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

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

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

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

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

$\frac{\sin ^{2}63^{o}+\sin ^{2}27^{o}}{\cos ^{2}17^{o}+\cos ^{2}73^{o}}$ ....................(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 $\sin^2\theta +\cos^2\theta = 1$ )

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

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

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

Therefore,

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

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

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

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, $9\times 1 = 9$

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

(A) 0 (B) 1 (C) 2 (D) –1

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

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

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

The correct option is (D)

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

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

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

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,

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

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

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.

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

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.

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

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

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.

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

[ Hint : Simplify LHS and RHS separately]

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.

$(v) \frac{\cos A-\sin A+1}{\cos A+\sin A-1}= \csc A+\cot A$ , using the identity $\csc ^{2}A= 1+\cot ^{2}A$

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.

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

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.

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.

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

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

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

[ Hint : Simplify LHS and RHS separately]

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.

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

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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### Questions related to CBSE Class 10th

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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.

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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!

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.

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