# 33 - Trigonometry Questions Answers

## (iii) tan69∘+tan66∘+1=tan69∘⋅tan66∘.

**Asked By: KHUSHI**

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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