120 - Mathematics Questions Answers

Solve

tanx-3cotx=2tan3x(0<x<360)

cosx-sinx=cosy-siny

cos2xcotx+1=cos2x+cotx

Asked By: YUMESH
is this question helpfull: 6 0 submit your answer
Joshi sir comment

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)

Asked By: ADARSH
is this question helpfull: 4 1 read solutions ( 2 ) | submit your answer
Joshi sir comment

use formula of tan(A+B)

Find the max. and min. value of

y = 7cosA + 24sinA

Asked By: ADARSH
is this question helpfull: 1 0 read solutions ( 3 ) | submit your answer
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

Asked By: ADARSH
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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

Asked By: ADARSH
is this question helpfull: 1 0 submit your answer
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

Asked By: JIGYASA KUMAR
is this question helpfull: 13 6 submit your answer
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 ?

Asked By: SIDDHANT
is this question helpfull: 4 0 submit your answer
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)

Asked By: VISHVESH CHATURVEDI
is this question helpfull: 4 1 submit your answer
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 

 

Asked By: RAMACHANDRAM
is this question helpfull: 2 1 submit your answer
Joshi sir comment

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

2cos7x/cos(3)+sin(3)>2^cos2x
Asked By: AARSHI
is this question helpfull: 10 0 submit your answer
Joshi sir comment

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

it is cos3 or cos3x 

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