135 - Mathematics Questions Answers

What is the value of 0⁄0? if its otherthan 1, then how this can be true:      lim      (ex-1)/x = 1

                                                                                                                                x −−>0

Asked By: BONEY HAVELIWALA
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Joshi sir comment

exact 0 and lim tends to 0 are different

Maths >> Trigonometry >> Complex Numbers Engineering Exam

what is the angular difference between the +j and -j ?

OR 

+4j and -4j  ?

 

Asked By: ADARSH KUMAR
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Joshi sir comment

180 degree

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6) = ?

Asked By: BONEY HAVELIWALA
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Joshi sir comment

sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)

= (π-2) + (π-4) + (6-2π) 

= 0

Maths >> Calculus >> Functions AIEEE

if ƒ:R−{0} -> R,  2ƒ(x) − 3ƒ(1⁄x) = x²  then ƒ(3)= ?

Asked By: BONEY HAVELIWALA
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Joshi sir comment

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)  

Maths >> Calculus >> Functions Others

Sir/Madam,

                    Suddenly a question struck on my mind:   0×∞= 1 or 0?? as 1⁄0=∞.

     

Asked By: BONEY HAVELIWALA
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Joshi sir comment

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

Asked By: YUMESH
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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
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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

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