166 - Questions Answers
∫√tanx
let √tanx = t
so sec2xdx/2√tanx = dt
so (1+t4)dx/2t = dt
so dx = 2tdt/(1+t4)
now solve
What is the value of 0⁄0? if its otherthan 1, then how this can be true: lim (ex-1)/x = 1
x −−>0
exact 0 and lim tends to 0 are different
what is the angular difference between the +j and -j ?
OR
+4j and -4j ?
180 degree
sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6) = ?
sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)
= (π-2) + (π-4) + (6-2π)
= 0
if ƒ:R−{0} -> R, 2ƒ(x) − 3ƒ(1⁄x) = x² then ƒ(3)= ?
2ƒ(x) − 3ƒ(1⁄x) = x² (1)
so 2ƒ(1/x) − 3ƒ(x) = 1/x² (2)
multiply (1) to 2 and (2) to 3, then add the two equations, you will get f(x) then calculate f(3)
Sir/Madam,
Suddenly a question struck on my mind: 0×∞= 1 or 0?? as 1⁄0=∞.
it will be 0, assume it by considering that ∞ is a big number
for ex. 0*(1111111111111111111111111) = 0
Solve
tanx-3cotx=2tan3x(0<x<360)
cosx-sinx=cosy-siny
cos2xcotx+1=cos2x+cotx
tanx - 3/tanx = 2[3tanx-tan3x]/[1-3tan2x]
so [tan2x - 3]/tanx = 2tanx[3-tan2x]/[1-3tan2x]
so [tan2x - 3][1-3tan2x] = 2tan2x[3-tan2x]
so [tan2x - 3][1-3tan2x] + 2tan2x[tan2x-3] = 0
so [tan2x - 3][1-tan2x] = 0
so tanx = 1, -1, √3, -√3
now check the values satisfying last eq.
then put these values in 2nd eq. to get answer.
Find the value :-
(tan69 +tan66) / (1 - tan69.tan66)
use formula of tan(A+B)
Find the max. and min. value of
y = 7cosA + 24sinA
7cosA + 24sinA
= 25[7cosA/25 + 24sinA/25] = 25[cosAcosB + sinAsinB] where tanB = 24/7
= 25cos[A-B] max and min of cos[A-B] = 1 and -1
so max. = 25 and min. = -25
Prove that :-
tanA + 2tan2A + 4tan4A + 8cot8A = cosecA / secA
We have to prove
tanA + 2tan2A + 4tan4A + 8cot8A = cosecA / secA
or tanA + 2tan2A + 4tan4A + 8cot8A = cotA
or 2tan2A + 4tan4A + 8cot8A = cotA - tanA
or tan2A + 2tan4A + 4cot8A = [1-tan2A]/2tanA
or tan2A + 2tan4A + 4cot8A = cot2A
or 2tan4A + 4cot8A = cot2A - tan2A
or tan4A + 2cot8A = [1-tan22A]/2tan2A
or tan4A + 2cot8A = cot4A
or 2cot8A = cot4A-tan4A
or cot8A = [1-tan24A]/2tan4A
it is true so proved