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.
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.
Express cosθ into all other trigonometric ratios.
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.
Determining the value of secθ in terms of tanθ.
1 + tan²θ = sec²θ
Substituting the value of secθ in (1)
Transformation into cotθ
We have got the cosθ in terms of tanθ
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
Substituting the value of sinθ in (1)
(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
(4)Hint: Check the relationship between cosecθ to other trigonometric ratios one by one.
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.
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
Find the value of tanθ if
sinθ = 7 : 25
Let perpendicular= 7x and hypotenuse= 25x
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
(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
Substituting sin²θ = 1 – cos²θ
= 2cosecθ= R.H.S
(5) Prove that ; (sinα + cosα)(tanα + cotα) = (secα + cosecα)
(sinα + cosα)(tanα + cotα)
= cosecα + secα= R.H.S
(6) Prove that
(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
(8) Prove that:
= 2secA = R.H.S
=1 + secθ.cosecθ= R.H.S
(10) Prove that:
= secA + tanA = R.H.S
(11) Prov that:
NCERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions of class 9 maths
|Chapter 1- Number System||Chapter 9-Areas of parallelogram and triangles|
|Chapter 2-Polynomial||Chapter 10-Circles|
|Chapter 3- Coordinate Geometry||Chapter 11-Construction|
|Chapter 4- Linear equations in two variables||Chapter 12-Heron’s Formula|
|Chapter 5- Introduction to Euclid’s Geometry||Chapter 13-Surface Areas and Volumes|
|Chapter 6-Lines and Angles||Chapter 14-Statistics|
|Chapter 7-Triangles||Chapter 15-Probability|
|Chapter 8- Quadrilateral|
NCERT Solutions of class 9 science
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NCERT solutions of class 10 science
Solutions of class 10 last years Science question papers
NCERT solutions of class 11 maths
|Chapter 1-Sets||Chapter 9-Sequences and Series|
|Chapter 2- Relations and functions||Chapter 10- Straight Lines|
|Chapter 3- Trigonometry||Chapter 11-Conic Sections|
|Chapter 4-Principle of mathematical induction||Chapter 12-Introduction to three Dimensional Geometry|
|Chapter 5-Complex numbers||Chapter 13- Limits and Derivatives|
|Chapter 6- Linear Inequalities||Chapter 14-Mathematical Reasoning|
|Chapter 7- Permutations and Combinations||Chapter 15- Statistics|
|Chapter 8- Binomial Theorem||Chapter 16- Probability|
NCERT solutions of class 12 maths
|Chapter 1-Relations and Functions||Chapter 9-Differential Equations|
|Chapter 2-Inverse Trigonometric Functions||Chapter 10-Vector Algebra|
|Chapter 3-Matrices||Chapter 11 – Three Dimensional Geometry|
|Chapter 4-Determinants||Chapter 12-Linear Programming|
|Chapter 5- Continuity and Differentiability||Chapter 13-Probability|
|Chapter 6- Application of Derivation||CBSE Class 12- Question paper of maths 2021 with solutions|
|Chapter 7- Integrals|
|Chapter 8-Application of Integrals|