हिंदी

Prove That: Tan 36° + Tan 9° + Tan 36° Tan 9° = 1 - Mathematics

Advertisements
Advertisements

प्रश्न

Prove that:
tan 36° + tan 9° + tan 36° tan 9° = 1

टिप्पणी लिखिए
Advertisements

उत्तर

\[\text{ We know that }36^\circ + 9^\circ = 45^\circ\]
Therefore, 
\[ \tan\left( 36^\circ + 9^\circ \right) = \tan45^\circ\]
\[ \Rightarrow \frac{\tan36^\circ + \tan9^\circ}{1 - \tan36^\circ \tan9^\circ} = 1\]
\[ \Rightarrow \tan36^\circ + \tan9^\circ = 1 - \tan36^\circ \tan9^\circ\]
\[ \Rightarrow \tan36^\circ + \tan9^\circ + \tan36^\circ \tan9^\circ = 1\]
Hence proved.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ २०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 17.3 | पृष्ठ २०

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Prove the following:

sin2 6x – sin2 4x = sin 2x sin 10x


Prove the following:

`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`


Prove the following:

cos 4x = 1 – 8sinx cosx


If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following:
sin (A − B)


Evaluate the following:
sin 36° cos 9° + cos 36° sin 9°


Evaluate the following:
 cos 80° cos 20° + sin 80° sin 20°


Prove that:
\[\frac{7\pi}{12} + \cos\frac{\pi}{12} = \sin\frac{5\pi}{12} - \sin\frac{\pi}{12}\]


Prove that

\[\frac{\cos 9^\circ + \sin 9^\circ}{\cos 9^\circ - \sin 9^\circ} = \tan 54^\circ\]

Prove that

\[\frac{\cos 8^\circ - \sin 8^\circ}{\cos 8^\circ + \sin 8^\circ} = \tan 37^\circ\]

 If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].


Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]


Prove that:
\[\frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x\]


Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.

 

If tan A = x tan B, prove that
\[\frac{\sin \left( A - B \right)}{\sin \left( A + B \right)} = \frac{x - 1}{x + 1}\]


If α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6 cos x + 8 sin x = 9, find the value of sin (α + β).

 

If \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].


If α + β − γ = π and sin2 α +sin2 β − sin2 γ = λ sin α sin β cos γ, then write the value of λ. 


If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.


If tan (A + B) = p and tan (A − B) = q, then write the value of tan 2B


The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\] 


If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to


If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\] 


If tan θ1 tan θ2 = k, then \[\frac{\cos \left( \theta_1 - \theta_2 \right)}{\cos \left( \theta_1 + \theta_2 \right)} =\]


If A − B = π/4, then (1 + tan A) (1 − tan B) is equal to 


Express the following as the sum or difference of sines and cosines:
2 sin 4x sin 3x


If sin(θ + α) = a and sin(θ + β) = b, then prove that cos 2(α - β) - 4ab cos(α - β) = 1 - 2a2 - 2b2

[Hint: Express cos(α - β) = cos((θ + α) - (θ + β))]


Find the general solution of the equation `(sqrt(3) - 1) costheta + (sqrt(3) + 1) sin theta` = 2

[Hint: Put `sqrt(3) - 1` = r sinα, `sqrt(3) + 1` = r cosα which gives tanα = `tan(pi/4 - pi/6)` α = `pi/12`]


If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to ______.


If f(x) = cos2x + sec2x, then ______.

[Hint: A.M ≥ G.M.]


The value of sin(45° + θ) - cos(45° - θ) is ______.


If sinθ + cosθ = 1, then the value of sin2θ is equal to ______.


If α + β = `pi/4`, then the value of (1 + tan α)(1 + tan β) is ______.


If tanα = `1/7`, tanβ = `1/3`, then cos2α is equal to ______.


If tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.


State whether the statement is True or False? Also give justification.

If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×