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Post a LessonAnswered on 23/12/2022 Learn CBSE/Class 12/Mathematics/Unit III: Calculus
Sonika Annad Academy
Answered on 01 Mar Learn CBSE/Class 12/Mathematics/Unit III: Calculus
Kalaiselvi
Online Mathematics tutor with 4 years experience(Online Classes for 10th to 12th)
Let sin-1(3/5) = x and sin-1(8/17) = y
Therefore sinx = 3/5 and siny = 8/17
Now, cosx = √(1 - sin2x) = √(1 - (3/5)2) = √(1 - 9/25) = 4/5 and cosy = √(1 - sin2y) = √(1 - (8/17)2) = √(1 - 64/289) = 15/17
We have cos(x - y) = cosx cosy + sinx siny = 4/5 x 15/17 + 3/5 x 8/17 = 60/85 + 24/85 = 84/85 ⇒ x - y = cos-1(84/85) ⇒ sin-1(3/5) - sin-1(8/17) = cos-1(84/85)
read lessAnswered on 01 Mar Learn CBSE/Class 12/Mathematics/Unit III: Calculus
Kalaiselvi
Online Mathematics tutor with 4 years experience(Online Classes for 10th to 12th)
Answered on 06 Apr Learn CBSE/Class 12/Mathematics/Unit III: Calculus
Sadika
Let's denote \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) \) as \( \alpha \) and \( \tan^{-1} \left( \frac{x + 1}{x + 2} \right) \) as \( eta \).
Given that \( \tan^{-1} \left( \frac{x - 1}{x - 2} \right) + \tan^{-1} \left( \frac{x + 1}{x + 2} \right) = \frac{\theta}{4} \), we can use the tangent addition formula:
\[ \tan(\alpha + eta) = \frac{\tan \alpha + \tan eta}{1 - \tan \alpha \cdot \tan eta} \]
Substitute \( \tan \alpha = \frac{x - 1}{x - 2} \) and \( \tan eta = \frac{x + 1}{x + 2} \):
\[ \tan(\alpha + eta) = \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \cdot \frac{x + 1}{x + 2}} \]
\[ \tan(\alpha + eta) = \frac{\frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)}}{1 - \frac{(x - 1)(x + 1)}{(x - 2)(x + 2)}} \]
\[ \tan(\alpha + eta) = \frac{x^2 + x - 2 + x^2 - x - 2}{(x - 2)(x + 2) - (x^2 - 1)} \]
\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{x^2 + 4 - x^2 + 1} \]
\[ \tan(\alpha + eta) = \frac{2x^2 - 4}{5} \]
Given that \( \tan(\alpha + eta) = \frac{\theta}{4} \), we have:
\[ \frac{2x^2 - 4}{5} = \frac{\theta}{4} \]
\[ 8x^2 - 16 = 5\theta \]
\[ 8x^2 = 5\theta + 16 \]
\[ x^2 = \frac{5\theta + 16}{8} \]
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]
So, the value of \( x \) depends on the value of \( \theta \).
In LaTeX code:
\[ x = \pm \sqrt{\frac{5\theta + 16}{8}} \]
Answered on 06 Apr Learn CBSE/Class 12/Mathematics/Unit III: Calculus
Sadika
To find the principal value of tan−1(1)tan−1(1), we need to determine the angle whose tangent is equal to 1.
Since tangent is defined as the ratio of the opposite side to the adjacent side in a right triangle, we can consider a right triangle where the angle whose tangent is 1 is one of its acute angles.
In a right triangle, if the ratio of the opposite side to the adjacent side is 1, then the opposite side and the adjacent side are equal in length. Therefore, we have a triangle with legs of equal length.
The angle whose tangent is 1 corresponds to a 45-degree angle (or π44π radians) in standard position.
So, the principal value of tan−1(1)tan−1(1) is π44π radians.
In LaTeX code: tan−1(1)=π4tan−1(1)=4π
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