Question

If tan 9°=p/q then find
sec^2 81/
1+cot^2 81
​

Answer 1

Answer:

P^2/p^2

Step-by-step explanation:

tan9=p/q

Sec^2 81/1+Cot^2

(p/q)^2

Answer 2

The value of the given expression is $(\dfrac{q}{p})^2$ .

Explanation:

Given : $\tan 9^{\circ}=\dfrac{p}{q}$

To find : $\dfrac{\sec^281^{\circ}}{1+\cot^281^{\circ}}$

Consider the given expression : $\dfrac{\sec^281^{\circ}}{1+\cot^281^{\circ}}$

$=\dfrac{\sec^281^{\circ}}{\csc^281^{\circ}}$  [$\because \ csc^2 x=1+cot^2 x$]

$=\dfrac{\dfrac{1}{\cos^2\ 81^{\circ}}}{\dfrac{1}{\sin^281^{\circ}}}=\dfrac{\sin^281^{\circ}}{\cos^2\ 81^{\circ}}$

$=\tan^2 81^{\circ}=( \cot (90^{\circ}-81^{\circ}))^2$ [∵ tan A= cot (90°-A) ]

$=(\cot9^{\circ})^2=(\dfrac{1}{\tan 9^{\circ}})^2$  

$=(\dfrac{1}{\dfrac{p}{q}})^2=(\dfrac{q}{p})^2$

Hence, the value of the given expression is $(\dfrac{q}{p})^2$ .

# Learn more :

Tan 9° *tan 27° * tan 63° * tan 81° =​

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