# 134 - Mathematics Questions Answers

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

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)

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-tan^{3}x]/[1-3tan^{2}x]

so [tan^{2}x - 3]/tanx = 2tanx[3-tan^{2}x]/[1-3tan^{2}x]

so [tan^{2}x - 3][1-3tan^{2}x] = 2tan^{2}x[3-tan^{2}x]

so [tan^{2}x - 3][1-3tan^{2}x] + 2tan^{2}x[tan^{2}x-3] = 0

so [tan^{2}x - 3][1-tan^{2}x] = 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**

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**Joshi sir comment**

use formula of tan(A+B)

Find the max. and min. value of

y = 7cosA + 24sinA

**Asked By: ADARSH**

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**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

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-tan^{2}A]/2tanA

or tan2A + 2tan4A + 4cot8A = cot2A

or 2tan4A + 4cot8A = cot2A - tan2A

or tan4A + 2cot8A = [1-tan^{2}2A]/2tan2A

or tan4A + 2cot8A = cot4A

or 2cot8A = cot4A-tan4A

or cot8A = [1-tan^{2}4A]/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

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

**Asked By: JIGYASA KUMAR**

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**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