NCERT Solutions for Class 12 Maths Chapter 2
NCERT Solutions for class 12 Maths Chapter 2 Inverse Trigonometric Functions in Hindi Medium and English Medium PDF file format to free download along with NCERT Solutions Apps updated for new academic session 2020-2021.The Previous Years Papers are to know the type and pattern of the questions asked which are designed as per latest CBSE Syllabus for the current academic session.
NCERT Solutions for class 12 Maths Chapter 2
|Chapter 2:||Inverse Trigonometric Functions|
12th Maths Chapter 2 Solutions
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Class 12 Maths Chapter 2 Solutions in Hindi Medium
Class 12 Maths Chapter 2 Solutions in PDF Format
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Inverse Trigonometric Functions
We know that a function has inverse if and only if it is one – one and onto. There are many functions like trigonometric functions are not one – one or onto because these functions periodic so these are many one. In order to get inverse of these functions, we must restrict their domain and co-domain in such a way that they become one – one and onto and these restricted values are known as principle values.
There are some important theorems/ properties of inverse trigonometric functions. It is important to know that how to convert one function into the terms of others.
Complement functions are useful during the conversion of one functions into it complimentary functions.
During the application of addition formula of tan, we should know that whether the product of x and y is greater than 1 or not. If it is more than 1, use the above formula and then apply the following formula.
The study of trigonometric functions started in 2nd millennium BC but in India it flourished in the Gupta Period due to Aryabhata. During the middle age, the study of trigonometry continued in Islamic Maths. The modern form of trigonometry functions were being used since 17th century by famous mathematician Isaac Newton, James Stirling etc.
Important Questions on 12th Maths Chapter 2
then sin y =-1/2=-sin(π/6)=sin(-π/6)
We know that the range of the principal value branch of sin−1 is [-π/2,π/2] and sin(-π/6)=-1/2
Therefore, the principal value of sin^(-1) (-1/2) is -π/6.
RHS = 〖sin〗^(-1) (3x-4x^3 )
=〖sin〗^(-1) (3 sinθ-4〖sin〗^3 θ)
=〖sin〗^(-1) (sin3θ )
= 3〖sin〗^(-1) x
We know that the range of the principal value branch of tan−1 is (-π/2,π/2) and tan(-π/4)=-1
Therefore, the principal value of tan−1 (−1) is -π/4.
⇒〖tan〗^(-1) (2cosx/(1-〖cos〗^2 x))=〖tan〗^(-1) (2cosecx )
[as 2〖tan〗^(-1) x
=〖tan〗^(-1) 2x/(1-x^2 )]
⇒ 2cosx/(1-〖cos〗^2 x)=2cosecx
⇒ 2cosx/(〖sin〗^2 x)=2/sinx
⇒2 sin〖x.cos〖x=2〖sin〗^2 x〗 〗
⇒ 2 sin〖x.cos〖x-2〖sin〗^2 x=0〗 〗
⇒2 sin〖x(cos〖x-sinx 〗 )=0〗
⇒ 2 sin〖x=0 or cos〖x-sinx 〗=0〗
But sinx≠0 as it does not satisfy the equation
⇒(〖sin〗^(-1) 1/5+〖cos〗^(-1) x)=〖sin〗^(-1) 1
⇒(〖sin〗^(-1) 1/5+〖cos〗^(-1) x)=π/2
⇒〖sin〗^(-1) 1/5 =〖sin〗^(-1) x
[∵ 〖sin〗^(-1) x+〖cos〗^(-1) x=π/2]