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NCERT Exemplar · Q64

Q.In a triangle ABCABC with ∠C=90∘\angle C = 90^\circ the equation whose roots are tan⁡A\tan A and tan⁡B\tan B is ______.

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In right triangle ABCABC with ∠C=90∘\angle C = 90^\circ, tan⁡A\tan A and tan⁡B\tan B are the ratios of the two legs to each other, so their product is always 11 and their sum equals c2ab\dfrac{c^2}{ab} by the Pythagorean theorem — giving the quadratic abx2−c2x+ab=0abx^2 - c^2x + ab = 0.

Let a=BCa = BC (the side opposite ∠A\angle A), b=CAb = CA (the side opposite ∠B\angle B), and c=ABc = AB (the hypotenuse, opposite the right angle at CC).

Step 1 — Write tan⁡A\tan A and tan⁡B\tan B in terms of the sides.

In the right triangle at CC:

tan⁡A=side opposite Aside adjacent to A=ab,tan⁡B=side opposite Bside adjacent to B=ba\tan A = \frac{\text{side opposite } A}{\text{side adjacent to } A} = \frac{a}{b}, \qquad \tan B = \frac{\text{side opposite } B}{\text{side adjacent to } B} = \frac{b}{a}

Step 2 — Find the product of the roots.

tan⁡A⋅tan⁡B=ab⋅ba=1\tan A \cdot \tan B = \frac{a}{b} \cdot \frac{b}{a} = 1

This product is always 11, regardless of the specific triangle — a direct consequence of AA and BB being complementary (A+B=90∘A+B=90^\circ, so tan⁡B=cot⁡A=1/tan⁡A\tan B = \cot A = 1/\tan A).

Step 3 — Find the sum of the roots.

tan⁡A+tan⁡B=ab+ba=a2+b2ab\tan A + \tan B = \frac{a}{b} + \frac{b}{a} = \frac{a^2+b^2}{ab} …

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