# 28 - Trigonometry 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

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

((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