If tan200 =λ, then tan1600 -tan11001+(tan1600)(tan1100)= - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{1 + λ^{2}}{2 λ}$
B
$\frac{1 + λ^{2}}{λ}$
C
$\frac{1 - λ^{2}}{λ}$
D
$\frac{1 - λ^{2}}{2 λ}$

Solution:

Give $\tan 20 ° = λ,$

Consider the expression $\frac{\tan 160^{0} - \tan 110^{0}}{1 + (\tan 160^{0}) (\tan 110^{0})}$

use the allied angles concept

$\begin{array}{rcl} \tan (160 °) & = & \tan (180 ° - 20 °) \\ = & - \tan 20 ° \\ = & - λ \end{array}$

and

$\begin{array}{rcl} \tan 110 ° & = & \tan (90 ° + 20 °) \\ = & - c o t 20 ° \\ = & - \frac{1}{λ} \end{array}$

Hence,

$\begin{array}{rcl} \frac{\tan 160^{0} - \tan 110^{0}}{1 + (\tan 160^{0}) (\tan 110^{0})} & = & \frac{- λ + \frac{1}{λ}}{1 + 1} \\ = & \frac{- λ^{2} + 1}{2 λ} \end{array}$

If $\tan 20^{0} = λ,$ then $\frac{\tan 160^{0} - \tan 110^{0}}{1 + (\tan 160^{0}) (\tan 110^{0})} =$

Solution:

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