Question

If cot 20° =P
Then tan 160°-tan110°/1+tan160+tan110

Answer 1

Answer:-

Given:-

cot 20° = P -- equation (1)

We know that,

cot θ = 1/tan θ

So,

⟹ 1/tan 20° = p

⟹ tan 20° = 1/p -- equation (2)

We have to find:-

(tan 160° - tan 110°) / (1 + tan 160° + tan 110°)

• tan 160° = tan (180° - 20°) = - tan 20° [ ∵ tan (180° - θ) ]
• tan 110° = tan (90° + 20°) = - cot 20° [ ∵ tan (90° + θ) = - cot θ ]

Putting the respective values we get,

⟹ [ - tan 20° - (- cot 20°) ] / 1 + ( - tan 20°) + ( - cot 20°)

⟹ [ - tan 20° + cot 20° ] / 1 - tan 20° - cot 20°

Substitute the values of tan 20° and cot 20° from equations (1) & (2).

⟹ ( - 1/p + p) / (1 - 1/p - p)

⟹ { ( - 1 + p ²) / p } * { p / (p - 1 - p²) }

⟹ (p² - 1) / (p - p² - 1)

Answer 2

Answer :

( p² - 1 ) / ( p - p² - 1 )

Explanation :

Before solving the problem, we must know some basic trigonometric formulas and identities which are very helpful for solving these types of problems.

Basic formulas to be used in solution,

• tan θ = 1/cot θ
• cot θ = 1/tan θ
• tan ( 90° + θ ) = - cot θ
• tan ( 180° - θ ) = - tan θ
• tan ( 90° - θ ) = cot θ

Now let's try to solve this problem, by converting the given terms in the form of these identities.

Firstly we are given,

⇒ cot 20° = p

⇒ tan θ = 1 / cot θ

⇒ tan 20° = 1 / cot 20°

⇒ tan 20° = 1 / p

We have to simplify,

⇒ ( tan 160° - tan 110° ) / ( 1 + tan 160° + tan 110° )

⇒ [ tan(180°-20°) - tan(90°+20°)] / [ 1 + tan(180°-20°) + tan(90°+20°)]

Now use the identities we discussed above,

⇒ [ - tan 20° - ( - cot 20° ) ] / [ 1 + (-tan 20°) + ( - cot 20°) ]

⇒ [ - tan 20° + cot 20° ] / [ 1 - tan 20° - cot 20° ]

Now substitute the given values

⇒ [ -1/p + p ] / [ 1 - 1/p - p ]

⇒ [ (-1 + p²)/p ] / [ ( p - 1 - p²)/p ]

⇒ (- 1 + p² ) / ( p - 1 - p²)

⇒ ( p² - 1 ) / ( p - p² - 1 )

Hence this is the required answer.

[ If you find any complexity in any step, kindly see the attachment ].