NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry
NCERT Solutions of important questions from Class 10 Maths Chapter 8 Introduction To Trigonometry are given here straightforward and well-ordered clarifications. These answers for Introduction To Trigonometry are amazingly well known among Class 10 students for Math Introduction To Trigonometry Solutions convenient for rapidly finishing your homework and getting ready for tests. All questions and answers from the NCERT Book of Class 10 Maths Chapter 8 are given here to you . You will likewise adore the advertisement-free understanding on future study point’s NCERT Solutions of Important questions. All NCERT Solutions of Important Questions for Class 10 Maths are set up by specialists and are 100% precise.
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NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry
Exercise 8.1- Introduction to Trigonometry
Exercise 8.2 -Introduction to Trigonometry
Exercise 8.3-Introduction to Trigonometry
NCERT solutions of Important Questions-Introduction to Trigonometry
NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry
For doing all the trigonometric solutions you should have the following formulas in your brain.
Point to be noticed that here perpendicular means side opposite to the angle and base means side adjacent to the angle, the hypotenuse is the longest side opposite to right angle as in the following fig.
, ,
NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry
Example –
Express cosθ into all other trigonometric ratios.
Solution-
transformation into sinθ
The relationship between cosθ and sinθ is given as
sin²θ + cos²θ = 1
cos²θ = 1 – sin²θ
Transformation into tanθ
There is no direct relation between cosθ and tanθ but we know the relationship between cosθ and secθ as following.
…………(1)
Determining the value of secθ in terms of tanθ.
1 + tan²θ = sec²θ
………(2)
Substituting the value of secθ in (1)
Transformation into cotθ
We have got the cosθ in terms of tanθ
………(1)
We know
………..(2)
Substituting this value of tanθ in (1)
Transformation into secθ
There is a direct relationship between cosθ and secθ
Transformation into cosecθ
While expressing cosθ into cosecθ, think over the following fact
………..(1)
Substituting the value of sinθ in (1)
NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry
Practice exercise-
(1) express sinθ in terms of tanθ.
(2) Write tan θ in phrases of cosecθ.
(3) decide the relationship between secθ and sinθ.
(4) remodel cosecθ into all trigonometric ratios.
(5)express cotθ in terms of secθ.
(1) Hint: sinθ⇒cosecθ⇒cotθ⇒tanθ
(2)Hint:tanθ⇒cotθ⇒cosecθ
(3)Hint:secθ⇒cosθ⇒sinθ
(4)Hint: Check the relationship between cosecθ to other trigonometric ratios one by one.
(5) Hint:cotθ⇒tanθ⇒secθ
Trigonometric Ratios of Complementary Angles-
∠A + ∠B + ∠C = 18O°
∠A + 90° + ∠C =180°
∠A + ∠C = 180°– 90°
∠A + ∠C = 90°
∠A & ∠C are complementry to each other
∠C = 90 –∠A
So, comparing the trigonometric ratios corresponding to ∠A, and ∠(90 –A)i.e∠C.
……………(1)
……(2)
From the equations (1) and (2) we have
sin(90–A) = cosA, cos (90– A) = sinA, tan(90–A) = cotA
cot(90–A) = tanA, sec (90– A) = cosecA, cosec(90–A) = sinA
Determining the value of other trigonometric ratio when value of one of them is given-
Table of trigonometric functions for cbse 10 th class
Study NCERT solutions for class 10 maths exercise 8.1 Introduction to trigonometry
Example-
Find the value of tanθ if
.
sinθ = 7 : 25
Let perpendicular= 7x and hypotenuse= 25x
Ans.
NCERT Solutions of Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry
Practice exercise- Evaluate the following.
(3) In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Study A to Z NCERT 10 th class maths solution of lesson 9.1 ‘Some Application Of the Trigonometry’
67° + 23° = 90° and 55° +25° = 90°⇒23= 90 –67 and 25 =90–55, so substituting th values of 23 and 25 in the given expression
(sin²θ +cos²θ=1)
(2) If ∠A and ∠B are acute angles such that cosA = cosB then show that ∠A = ∠B.
Answer- Let ΔABC is a right triangle in which ∠A and ∠B are acute angles.
We are given that cosA = cosB
AC = BC
∠B = ∠A(angles to equal opposite sides)
See the video for solutions of Q5 part(i),part(ii) and part(iii) of exercise 8.4 class 10 NCERT maths
L.H.S
Substituting sin²θ = 1 – cos²θ
= R.H.S
Solution-
L.H.S
= 2cosecθ= R.H.S
(5) Prove that ; (sinα + cosα)(tanα + cotα) = (secα + cosecα)
Solution-
L.H.S
(sinα + cosα)(tanα + cotα)
= cosecα + secα= R.H.S
(6) Prove that
Solution-
L.H.S
= R.H.S
(7) Prove that
Since in the R.H.S we have to get secθ and tanθ so transforming L.H.S in terms of secθ and tanθ, dividing the numerator and denominator by cosθ, we get
Let the [tanθ – secθ] remain in its place at the denominator since in the R.H.S we are to be required this one.
So, multiplying the expressions in the numerators
= R.H.S
(8) Prove that:
L.H.S
= 2secA = R.H.S
(9)
Solution-
L.H.S
=1 + secθ.cosecθ= R.H.S
(10) Prove that:
L.H.S
= secA + tanA = R.H.S
(11) Prove that:
L.H.S
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