161 - Mathematics Questions Answers

it will be 0, assume it by considering that ∞ is a big number 

for ex.   0*(1111111111111111111111111) = 0

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.

use formula of tan(A+B)

   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

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

sin(A+B+C+D) = sin(A+B)[cosCcosD-sinCsinD]+cos(A+B)[sinCcosD+cosCsinD]

similarly solve all the terms and add all these terms for getting answer

OPTION

ordered way INOOPT

total words = 6!/2! = 360

so rank = (360/6)*2+([60*2]/5)*3+(24/4)*3+(6/3)*0+(2/2)*1+1 = 120+72+18+0+1+1 = 212

f(x)= log₄log₃log₂x 

so 0 < log₃log₂x < ∞

so 1 < log₂x < ∞ 

so 2 < x < ∞

Submit the correct question, In first bracket / will be there so

((1/(secA.secA-cosA.cosA))-(1/(cosecA.cosecA-sinA.sinA))).sinA.sinA.cosA.cosA

= [cos²A/sin²A(1+cos²A) - sin²A/cos²A(1+sin²A)]sin²Acos²A

= [cos⁴A(1+sin²A) - sin⁴A(1+cos²A)] / [(1+sin²A)(1+cos²A)]

now solve

x²+1 > 2x so cosA>1 which is not possible