मराठी

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

Advertisements
Advertisements

प्रश्न

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

रिकाम्या जागा भरा
Advertisements

उत्तर

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometric Functions - Exercise [पृष्ठ ५९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 3 Trigonometric Functions
Exercise | Q 64 | पृष्ठ ५९

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the principal and general solutions of the equation `tan x = sqrt3`


Find the principal and general solutions of the equation  `cot x = -sqrt3`


Find the general solution of the equation cos 4 x = cos 2 x


If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that

\[\frac{1 - \cos x + \sin x}{1 + \sin x}\] is also equal to a.

If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]


If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]


Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]


Prove that: tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0


Prove that

\[\frac{\sin(180^\circ + x) \cos(90^\circ + x) \tan(270^\circ - x) \cot(360^\circ - x)}{\sin(360^\circ - x) \cos(360^\circ + x) cosec( - x) \sin(270^\circ + x)} = 1\]

 


Prove that

\[\left\{ 1 + \cot x - \sec\left( \frac{\pi}{2} + x \right) \right\}\left\{ 1 + \cot x + \sec\left( \frac{\pi}{2} + x \right) \right\} = 2\cot x\]

 


Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]

 

Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]


\[\sqrt{\frac{1 + \cos x}{1 - \cos x}}\] is equal to

 


If \[0 < x < \frac{\pi}{2}\], and if \[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\], then y is equal to


If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to


If tan A + cot A = 4, then tan4 A + cot4 A is equal to


If tan θ + sec θ =ex, then cos θ equals


Find the general solution of the following equation:

\[\sin x = \frac{1}{2}\]

Find the general solution of the following equation:

\[\tan x = - \frac{1}{\sqrt{3}}\]

Find the general solution of the following equation:

\[\cos 3x = \frac{1}{2}\]

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equation:

\[\cos x + \cos 2x + \cos 3x = 0\]

Solve the following equation:

\[\sin 3x - \sin x = 4 \cos^2 x - 2\]

Solve the following equation:
\[\cot x + \tan x = 2\]

 


Write the general solutions of tan2 2x = 1.

 

Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]

 and cos 2x are in A.P.


If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.


If \[4 \sin^2 x = 1\], then the values of x are

 


If \[\cos x = - \frac{1}{2}\] and 0 < x < 2\pi, then the solutions are


The number of values of x in the interval [0, 5 π] satisfying the equation \[3 \sin^2 x - 7 \sin x + 2 = 0\] is


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

2 sin2x + 1 = 3 sin x


Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0


Choose the correct alternative:
If tan α and tan β are the roots of x2 + ax + b = 0 then `(sin(alpha + beta))/(sin alpha sin beta)` is equal to


Choose the correct alternative:
If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval


Choose the correct alternative:
`(cos 6x + 6 cos 4x + 15cos x + 10)/(cos 5x + 5cs 3x + 10 cos x)` is equal to


Solve the equation sin θ + sin 3θ + sin 5θ = 0


Solve `sqrt(3)` cos θ + sin θ = `sqrt(2)`


If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×