# 158 - Mathematics Questions Answers

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

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

**Asked By: SIDDHANT**

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**Joshi sir comment**

f(x)= log_{4}log_{3}log_{2}x

so 0 < log_{3}log_{2}x < ∞

so 1 < log_{2}x < ∞

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**

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**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

= [cos^{2}A/sin^{2}A(1+cos^{2}A) - sin^{2}A/cos^{2}A(1+sin^{2}A)]sin^{2}Acos^{2}A

= [cos^{4}A(1+sin^{2}A) - sin^{4}A(1+cos^{2}A)] / [(1+sin^{2}A)(1+cos^{2}A)]

now solve

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

**Asked By: RAMACHANDRAM**

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**Joshi sir comment**

x^{2}+1 > 2x so cosA>1 which is not possible

**Asked By: AARSHI**

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**Joshi sir comment**

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

it is cos3 or cos3x

**Asked By: AARSHI**

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**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

tan^{2}x tan^{2}3x tan4x = tan^{2}x - tan^{2}3x + tan4x

so tan4x = [tan^{2}x - tan^{2}3x] / [tan^{2}x tan^{2}3x-1]

now split RHS terms by formula x^{2}-y^{2}= (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

**Asked By: GAUTHAM GANESH**

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**Joshi sir comment**

by diagram (6,4),(6,3),(6,2) are points having AB<4 besides it two upside points are also there by symmetry not shown in diagram. These points are (6,5) and (6,6)

Total no. of points = 5