Advertisements
Advertisements
प्रश्न
If sinx + cosx = a, then |sinx – cosx| = ______.
उत्तर
Given that: sinx + cosx = a
(sinx + cosx)2 = a2
⇒ sin2x + cos2x + 2sinx cosx = a2
⇒ 1 + 2sinx cosx = a2
⇒ sinx cosx = `(a^2 - 1)/2` .......(i)
|sinx – cosx| = sin2x + cos2x – 2sinx cosx
= `1 - 2((a^2 - 1)/2)`
= 1 – (a2 – 1)
= 1 – a2 + 1
= 2 – a2
∴ |sinx – cosx| = `sqrt(2 - a^2)`
APPEARS IN
संबंधित प्रश्न
Prove the following:
cos2 2x – cos2 6x = sin 4x sin 8x
Prove the following:
`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`
Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x
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 \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], 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:
cos (A + B)
If \[\tan A = \frac{3}{4}, \cos B = \frac{9}{41}\], where π < A < \[\frac{3\pi}{2}\] and 0 < B <\[\frac{\pi}{2}\], find tan (A + B).
Prove that:
Prove that:
Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].
Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]
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
Prove that:
If sin α − sin β = a and cos α + cos β = b, then write the value of cos (α + β).
If tan \[\alpha = \frac{1}{1 + 2^{- x}}\] and \[\tan \beta = \frac{1}{1 + 2^{x + 1}}\] then write the value of α + β lying in the interval \[\left( 0, \frac{\pi}{2} \right)\]
If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =
If A + B + C = π, then \[\frac{\tan A + \tan B + \tan C}{\tan A \tan B \tan C}\] is equal to
If \[\cos P = \frac{1}{7}\text{ and }\cos Q = \frac{13}{14}\], where P and Q both are acute angles. Then, the value of P − Q is
If cot (α + β) = 0, sin (α + 2β) is equal to
If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B 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 sin 4x sin 3x
Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β) = cos 2α
Match each item given under column C1 to its correct answer given under column C2.
| C1 | C2 |
| (a) `(1 - cosx)/sinx` | (i) `cot^2 x/2` |
| (b) `(1 + cosx)/(1 - cosx)` | (ii) `cot x/2` |
| (c) `(1 + cosx)/sinx` | (iii) `|cos x + sin x|` |
| (d) `sqrt(1 + sin 2x)` | (iv) `tan x/2` |
If tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.
State whether the statement is True or False? Also give justification.
If tanA = `(1 - cos B)/sinB`, then tan2A = tanB
In the following match each item given under the column C1 to its correct answer given under the column C2:
| Column A | Column B |
| (a) sin(x + y) sin(x – y) | (i) cos2x – sin2y |
| (b) cos (x + y) cos (x – y) | (ii) `(1 - tan theta)/(1 + tan theta)` |
| (c) `cot(pi/4 + theta)` | (iii) `(1 + tan theta)/(1 - tan theta)` |
| (d) `tan(pi/4 + theta)` | (iv) sin2x – sin2y |
