Advertisements
Advertisements
Question
Prove that:
tan 36° + tan 9° + tan 36° tan 9° = 1
Advertisements
Solution
\[\text{ We know that }36^\circ + 9^\circ = 45^\circ\]
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.
APPEARS IN
RELATED QUESTIONS
Prove that: `sin^2 pi/6 + cos^2 pi/3 - tan^2 pi/4 = -1/2`
Prove that `2 sin^2 pi/6 + cosec^2 (7pi)/6 cos^2 pi/3 = 3/2`
Find the value of: tan 15°
Prove the following:
sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Prove the following:
`cos ((3pi)/4 + x) - cos((3pi)/4 - x) = -sqrt2 sin x`
Prove the following:
`(sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x`
Prove the following:
`tan 4x = (4tan x(1 - tan^2 x))/(1 - 6tan^2 x + tan^4 x)`
Prove that: `((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x)) = tan 6x`
If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
sin (A + B)
Evaluate the following:
sin 36° cos 9° + cos 36° sin 9°
If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
cos (A + B)
Prove that:
Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]
Prove that:
tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x
Prove that:
\[\tan\frac{\pi}{12} + \tan\frac{\pi}{6} + \tan\frac{\pi}{12}\tan\frac{\pi}{6} = 1\]
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 tan (A + B) = x and tan (A − B) = y, find the values of tan 2A and tan 2B.
If tan x + \[\tan \left( x + \frac{\pi}{3} \right) + \tan \left( x + \frac{2\pi}{3} \right) = 3\], then prove that \[\frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1\].
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 x cos θ = y cos \[\left( \theta + \frac{2\pi}{3} \right) = z \cos \left( \theta + \frac{4\pi}{3} \right)\]then write the value of \[\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\]
Write the maximum value of 12 sin x − 9 sin2 x.
Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies.
If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to
If \[\tan\theta = \frac{1}{2}\] and \[\tan\phi = \frac{1}{3}\], then the value of \[\tan\phi = \frac{1}{3}\] is
If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
If A − B = π/4, then (1 + tan A) (1 − tan B) is equal to
If `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].
If tanθ = `(sinalpha - cosalpha)/(sinalpha + cosalpha)`, then show that sinα + cosα = `sqrt(2)` cosθ.
[Hint: Express tanθ = `tan (alpha - pi/4) theta = alpha - pi/4`]
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 tanα = `m/(m + 1)`, tanβ = `1/(2m + 1)`, then α + β is equal to ______.
The value of tan3A - tan2A - tanA is equal to ______.
State whether the statement is True or False? Also give justification.
If tanA = `(1 - cos B)/sinB`, then tan2A = tanB
State whether the statement is True or False? Also give justification.
If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
