# 158 - Mathematics Questions Answers

Find the max. and min. value of

y = 7cosA + 24sinA

Joshi sir comment

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

Maths >> Trigonometry >> General Trigonometry Engineering Exam

Prove that :-

tanA + 2tan2A + 4tan4A + 8cot8A = cosecA / secA

Joshi sir comment

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

Prove that :-

sin(A+B+C+D) + sin(A+B-C-D) + sin(A+B-C+D) + sin(A+B+C-D) = 4sin(A+B).cosC.cosD

Joshi sir comment

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

what will be the rank of word OPTION in the dictionary of words formed by letters of the word OPTION

Joshi sir comment

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

Maths >> Calculus >> Functions IIT JEE

FIND DOMAIN OF f(x)=log4log3 log2x WHERE 4,3,2 ARE BASES ?

Joshi sir comment

f(x)= log4log3log2

so 0 < log3log2x <

so 1 < log2x < ∞

so 2 < x <

((1(secA.secA-cosA.cosA))-(1/(cosecA.cosecA-sinA.sinA))).sinA.sinA.cosA.cosA=(1-cosA.cosA.sinA.sinA)/(2+cosA.cosA.sinA.sinA)

Joshi sir comment

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

if x is any real number and cosA=x^2+1/2x then cos A value is

Joshi sir comment

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

2cos7x/cos(3)+sin(3)>2^cos2x
Joshi sir comment

Please submit what we have to do and also correct the question

it is cos3 or cos3x

(tanx)^2(tan3x)^2(tan4x)=(tanx)^2-(tan3x) ^2 + tan4x.
Joshi sir comment

I dont know what the question is but if the question is to solve this eq. then the method will be as given below

tan2x tan23x tan4x = tan2x  - tan23x + tan4x

so tan4x =  [tan2x  - tan23x] / [tan2x tan23x-1]

now split RHS terms by formula x2-y2= (x+y)(x-y)

so RHS = tan4xtan2x

now solve

A(3,4) and B is a variable point on the line |X| = 6. Also, AB ≤ 4. Then the number of positions of B with integral co-ordinates is:

(a)5  (b) 6 (c) 10 (d) 12 