Question

Cot20=p then tan160° - tan 110° ) / ( 1 + tan 60° tan 110)

Answer 1

Please see the attachment

Answer 2

Answer:

$\frac{tan 160 - tan110}{1+tan160\ tan110}$ =  $\frac{p^2 - 1}{2p}$

Step-by-step explanation:

Given,

cot 20 = p

Required to find

$\frac{tan 160 - tan110}{1+tan160\ tan110}$

Recall the formulas

1. tan(A -B) = $\frac{tan A - tan B}{1+tanAtanB}$
2. tan(90-A) = cot A
3. Cot 2A = $\frac{cot^2A - 1}{2cotA}$

Solution:

$\frac{tan 160 - tan110}{1+tan160\ tan110}$ is of the form  $\frac{tan A - tan B}{1+tanAtanB}$ where A = 160 and B = 110

Since $\frac{tan A - tan B}{1+tanAtanB}$ = tan(A-B), we have

$\frac{tan 160 - tan110}{1+tan160\ tan110}$ = tan(160 -110)  = tan 50

Again,

tan 50 = tan (90 -40) = cot 40

$\frac{tan 160 - tan110}{1+tan160\ tan110}$ = cot 40

We know, cot 2A = $\frac{cot^2A - 1}{2cotA}$

Substituting A = 20 we get

cot 40 = $\frac{cot^2 20 - 1}{2cot 20}$

Substituting the value of cot cot20 = p we get

cot 40 = $\frac{p^2 - 1}{2p}$

∴ $\frac{tan 160 - tan110}{1+tan160\ tan110}$ =  $\frac{p^2 - 1}{2p}$

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