Advertisements
Advertisements
Question
Prove that
Advertisements
Solution
\[\text{ LHS }= \frac{\cos8^\circ - \sin8^\circ}{\cos8^\circ + \sin8^\circ}\]
\[ = \frac{\frac{\cos8^\circ}{\cos8^\circ} - \frac{\sin8^\circ}{\cos8^\circ}}{\frac{\cos8}{\cos8} + \frac{\sin8}{\cos8}} \left( \text{ Dividing numeraor and denominator by }\cos 8^\circ \right)\]
\[ = \frac{1 - \tan8^\circ}{1 + \tan8^\circ}\]
\[ = \frac{1 - \tan8^\circ}{1 + 1 \times \tan8^\circ}\]
\[ = \frac{\tan45^\circ - \tan8^\circ}{1 + \tan45^\circ \tan8^\circ} \left( \text{ As }\tan 45^\circ = 1 \right)\]
\[ = \tan\left( 45^\circ - 8^\circ \right) \left[\text{ As }\frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan\left( A + B \right) \right]\]
\[ = \tan37^\circ\]
= RHS
Hence proved.
APPEARS IN
RELATED QUESTIONS
Prove that: `2 sin^2 (3pi)/4 + 2 cos^2 pi/4 + 2 sec^2 pi/3 = 10`
Prove the following: `cos (pi/4 xx x) cos (pi/4 - y) - sin (pi/4 - x)sin (pi/4 - y) = sin (x + y)`
Prove the following: `(tan(pi/4 + x))/(tan(pi/4 - x)) = ((1+ tan x)/(1- tan x))^2`
Prove the following:
`cos ((3pi)/ 2 + x ) cos(2pi + x) [cot ((3pi)/2 - x) + cot (2pi + x)]= 1`
Prove the following:
sin2 6x – sin2 4x = sin 2x sin 10x
Prove the following:
cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Prove that: `(cos x + cos y)^2 + (sin x - sin y )^2 = 4 cos^2 (x + y)/2`
Prove that: `((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x)) = tan 6x`
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:
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:
cos (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:
tan (A + B)
Prove that:
Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]
Prove that:
sin2 B = sin2 A + sin2 (A − B) − 2 sin A cos B sin (A − B)
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 + B) = x and tan (A − B) = y, find the values of tan 2A and tan 2B.
If cos A + sin B = m and sin A + cos B = n, prove that 2 sin (A + B) = m2 + n2 − 2.
If sin (α + β) = 1 and sin (α − β) \[= \frac{1}{2}\], where 0 ≤ α, \[\beta \leq \frac{\pi}{2}\], then find the values of tan (α + 2β) and tan (2α + β).
If sin α + sin β = a and cos α + cos β = b, show that
The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\]
If cot (α + β) = 0, sin (α + 2β) is equal to
If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B is
Express the following as the sum or difference of sines and cosines:
2 sin 4x sin 3x
If α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan (α + β) = `(2ac)/(a^2 - c^2)`.
If `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].
If cos(θ + Φ) = m cos(θ – Φ), then prove that 1 tan θ = `(1 - m)/(1 + m) cot phi`
[Hint: Express `(cos(theta + Φ))/(cos(theta - Φ)) = m/1` and apply Componendo and Dividendo]
If f(x) = cos2x + sec2x, then ______.
[Hint: A.M ≥ G.M.]
If tan θ = 3 and θ lies in third quadrant, then the value of sin θ ______.
If tanα = `m/(m + 1)`, tanβ = `1/(2m + 1)`, then α + β is equal to ______.
The value of tan3A - tan2A - tanA is equal to ______.
If sinθ + cosθ = 1, then the value of sin2θ is equal to ______.
If tanα = `1/7`, tanβ = `1/3`, then cos2α is equal to ______.
Given x > 0, the values of f(x) = `-3cos sqrt(3 + x + x^2)` lie in the interval ______.
The maximum distance of a point on the graph of the function y = `sqrt(3)` sinx + cosx from x-axis is ______.
State whether the statement is True or False? Also give justification.
If tanA = `(1 - cos B)/sinB`, then tan2A = tanB
