In a triangle ABC with &ang;C = 90° the equation whose roots are tan A and tan B is ______.

In a triangle ABC with &ang;C = 90° the equation whose roots are tan A and tan B is `underline(x^2 - (2/(sin 2A)) x + 1` = 0. Explanation: Given a ΔABC with &ang;C = 90° So, the equation whose roots are tanA and tanB is x2 – (tanA + tanB)x + tanA.tanB = 0 A + B = 90° ......[∵ &ang;C = 90°] ⇒ tan(A + B) = tan90° ⇒ `(tanA + tanB)/(1 - tanA tanB) = 1/0` ⇒ 1 – tanA tanB = 0 ⇒ tan A tan B = 1 .......(i) Now tanA + tanB = `sinA/cosA + sinB/cosB` = `(sinA cosB + cosA sinB)/(cosA cosB)` = `(sin(A + B))/(cosA cosB)` = `(sin 90^circ)/(cosA. cos(90^circ - A))` = `1/(cosA sinA)` &there4; tanA + tanB = `2/(2sinA cosA)` = `2/(sin 2A)` ......(ii) From (i) and (ii) we get `x^2 - (2/(sin 2A)) x + 1` = 0

In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______. - Mathematics

Question

In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.

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Solution

In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is `underline(x^2 - (2/(sin 2A)) x + 1` = 0.

Explanation:

Given a ΔABC with ∠C = 90°

So, the equation whose roots are tanA and tanB is

x² – (tanA + tanB)x + tanA.tanB = 0

A + B = 90° ......[∵ ∠C = 90°]

⇒ tan(A + B) = tan90°

⇒ `(tanA + tanB)/(1 - tanA tanB) = 1/0`

⇒ 1 – tanA tanB = 0

⇒ tan A tan B = 1 .......(i)

Now tanA + tanB = `sinA/cosA + sinB/cosB`

= `(sinA cosB + cosA sinB)/(cosA cosB)`

= `(sin(A + B))/(cosA cosB)`

= `(sin 90^circ)/(cosA. cos(90^circ - A))`

= `1/(cosA sinA)`

∴ tanA + tanB = `2/(2sinA cosA)`

= `2/(sin 2A)` ......(ii)

From (i) and (ii) we get

`x^2 - (2/(sin 2A)) x + 1` = 0

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Trigonometric Equations

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Q 64 Q 63.(ii)Q 65

Chapter 3: Trigonometric Functions - Exercise [Page 59]

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NCERT Exemplar Mathematics [English] Class 11

Chapter 3 Trigonometric Functions
Exercise | Q 64 | Page 59

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