Question

The tangents of two acute-angles aré 3 and 2. The sine of twice their difference is -
(A) 7/24
B) 7/48
(C) 7/50
(D) 7/25
​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Option C

7/50

Step-by-step explanation:

Given Problem:

The tangents of two acute-angles aré 3 and 2. The sine of twice their difference is -

(A) 7/24

B) 7/48

(C) 7/50

(D) 7/25

​Solution:

As slope of acute angle are 3 and 2 respectively.

So tanθθ = 3 and tan ϕϕ= 2

Hence,

The differnce of angles are:

$\sf tan^-^1 \bigg(3\bigg)-tan^-^1\bigg(2\bigg)=tan^-^1\bigg(\dfrac{3-2}{1+6}\bigg) = tan^-^1\bigg(\dfrac{1}{7}\bigg)\\\\ Let\: tan^-^1 \dfrac{1}{7}= y\\\ Hence,tan\:y=\dfrac{1}{7},Hence\: y =\dfrac{1}{\sqrt{50}}\\\ So,tan^-^1 \dfrac{1}{7} = sin^-^1 = \dfrac{1}{\sqrt{50}}\\\\\ Hence,The\:answer\:is\:7/50$