Advertisements
Advertisements
प्रश्न
If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to ______.
विकल्प
1
4
2
None of these
Advertisements
उत्तर
If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to 2.
Explanation:
sinθ + cosecθ = 2
Squaring L.H.S and R.H.S
We get,
⇒ (sinθ + cosecθ)2 = 22
⇒ (sinθ + cosecθ)2 = 4
⇒ sin2θ + cosec2θ + 2sinθ cosecθ = 4 [∵ `1/sintheta = cosectheta`]
= sin2θ + cosec2θ + 2 = 4
= sin2θ + cosec2θ = 2
APPEARS IN
संबंधित प्रश्न
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:
sin2 6x – sin2 4x = sin 2x sin 10x
Prove the following:
`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`
Prove the following:
`(sin x + sin 3x)/(cos x + cos 3x) = tan 2x`
Prove the following:
`tan 4x = (4tan x(1 - tan^2 x))/(1 - 6tan^2 x + tan^4 x)`
Prove that: `(cos x - cosy)^2 + (sin x - sin y)^2 = 4 sin^2 (x - y)/2`
Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
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)
If \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], where A and B both lie in second quadrant, find the value of sin (A + B).
If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
sin (A + B)
If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].
Prove that:
\[\tan\frac{\pi}{12} + \tan\frac{\pi}{6} + \tan\frac{\pi}{12}\tan\frac{\pi}{6} = 1\]
Prove that:
tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x
Prove that:
\[\frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x\]
If tan (A + B) = x and tan (A − B) = y, find the values of tan 2A and tan 2B.
If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) \[\frac{1}{a} - \frac{1}{b}\].
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 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
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 and β are two solutions of the equation a tan x + b sec x = c, then find the values of sin (α + β) and cos (α + β).
Find the maximum and minimum values of each of the following trigonometrical expression:
\[5 \cos x + 3 \sin \left( \frac{\pi}{6} - x \right) + 4\]
Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies.
If \[\frac{\cos \left( x - y \right)}{\cos \left( x + y \right)} = \frac{m}{n}\] then write the value of tan x tan y.
The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\]
If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =
tan 3A − tan 2A − tan A =
If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
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 cos 7x cos 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)`.
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`]
The value of tan3A - tan2A - tanA is equal to ______.
The value of `cot(pi/4 + theta)cot(pi/4 - theta)` is ______.
If α + β = `pi/4`, then the value of (1 + tan α)(1 + tan β) is ______.
State whether the statement is True or False? Also give justification.
If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
