Question

tan20°=lambda ,then tan160°-tan110°/1+(tan160°)(tan110°)=

Answer 1

Answer:

The value of expression is:

$\dfrac{\tan 160-\tan 110}{1+(\tan160)(\tan110)}=\dfrac{1-\lambda^2}{2\lambda}$

Step-by-step explanation:

We are given:

tan 20°=λ

Now we are asked to find the value of:

$\dfrac{\tan 160-\tan 110}{1+(\tan160)(\tan110)}$

We know that:

$\tan (180-\theta)=-\tan \theta$

and

$\tan (90+\theta)=-\cot \theta$

Hence, here our expression changes to:

Here we have:

$\tan 160=\tan (180-20)=-\tan 20$

similarly, we have:

$\tan 110=\tan (90+20)=-\cot 20$

Also we know that:

$\tan \theta=\dfrac{1}{\cot \theta}$

Hence,

$\cot 20=\dfrac{!}{\tan 20}\\\\\cot 20=\dfrac{1}{\lambda}$

Hence, we have:

$\dfrac{\tan 160-\tan 110}{1+(\tan160)(\tan110)}=\dfrac{-\tan 20-(-\cot \20)}{1+(-\tan 20)\times (-\cot 20)}\\\\\\=\dfrac{-\lambda+\dfrac{1}{\lambda}}{1+\lambda\times \dfrac{1}{\lambda}}\\\\\\=\dfrac{\dfrac{-(\lambda)^2+1}{\lambda}}{1+1}\\\\=\dfrac{1-\lambda^2}{2\lambda}$

Hence, the value of the expression is:

$\dfrac{\tan 160-\tan 110}{1+(\tan160)(\tan110)}=\dfrac{1-\lambda^2}{2\lambda}$

Answer 2

Answer:

1-p^2/2p

Step-by-step explanation:

LHS= tan 160-tan 110/1+tan 160 tan 110

=tan(160-110)=tan 50 = cot 40 = 1/tan 40

1/2tan 20/1-tan^2 20 = 1- p^2/2p RHS