If A + B + A + B = A B - A B and 0^{} [TG EAPCET 10 May 2024 Shift 1]

Correct Answer is : 135∘135^{}135∘

Correct Answer is : 135∘135^{}135∘

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TS EAMCET 10 May 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 23 of 160

Marks: +1, -0

\tan A + \tan B + \cot A + \cot B = \tan A \tan B - \cot A \cot B

0^{\circ} < A + B < 270^{\circ}

A + B =

Solution:

Given the problem:

\tan A + \tan B + \cot A + \cot B = \tan A \tan B - \cot A \cot B

We can express the trigonometric functions in terms of sine and cosine. That gives:

\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} + \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} = \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B} - \frac{\cos A}{\sin A} \cdot \frac{\cos B}{\sin B}

Simplifying each term, we get:

\frac{\sin^2 A + \cos^2 A}{\sin A \cos A} + \frac{\sin^2 B + \cos^2 B}{\sin B \cos B} = \frac{\sin^2 A \sin^2 B - \cos^2 A \cos^2 B}{\cos A \cos B \sin A \sin B}

Since

\sin^2 \theta + \cos^2 \theta = 1

, this reduces to:

\frac{1}{\sin A \cos A} + \frac{1}{\sin B \cos B} = \frac{\sin^2 A \sin^2 B - (1 - \sin^2 A)(1 - \sin^2 B)}{\cos A \cos B \sin A \sin B}

\Rightarrow \frac{1}{2}(\sin 2A + \sin 2B) = \sin^2 A + \sin^2 B - 1

Using the sine addition and subtraction formulas,

\sin(2x) = 2 \sin x \cos x

, transform the equation:

\sin(A+B) \cos(A-B) = \sin^2 A - \cos^2 B

Simplifying further:

\sin(A+B) \cos(A-B) = -\cos(A+B) \cos(A-B)

This implies:

\tan(A+B) = -1 \Rightarrow A+B = 135^{\circ}

Given the constraint

0^{\circ} < A+B < 270^{\circ}

, the valid solution is:

A+B = 135^{\circ}

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