TRIGONOMETRY PROVING PROBLEMS FOR CLASS 11

Problem 1 :

If A + B = 45°, show that

(1 + tan A)(1 + tan B)  =  2

Solution :

A + B  =  45°

tan (A + B)  =  tan 45°

(tan A + tan B)/(1 - tan A tan B)  =  1

tan A + tan B  =  1 - tan A tan B

Add tan A tan B on both sides.

tan A + tan B + tan A tan B =  1 

tan A + tan B(1 + tan A)  =  1 

Add 1 on both sides

1 + tan A + tan B(1 + tan A)  =  1 + 1

(1 + tan A) (1 + tan B)  =  2

Hence proved.

Problem 2 :

Prove that (1 + tan1°)(1 + tan2°)(1 + tan3°).......(1 + tan44°) is a multiple of 4. 

Solution :

=  (1 + tan1°)(1 + tan2°)(1 + tan3°)................(1 + tan44°)

=  (1 + tan1°)(1 + tan44°)(1 + tan2°)(1 + tan43°)(1 + tan3°)(1 + tan42°)................

If A + B  =  45°, then

(1 + tanA)(1 + tanB)  =  2

In the first two terms, we have 1 instead of A and 44 instead of B.

In a same manner, in third and 4th terms we have 2 instead of A and 43 instead of B. So, we get

  =  2(2)(2)........................

It is a multiple of 4.

Problem 3 :

Prove that

tan (π/4 + θ) - tan (π/4 − θ)  =  2 tan2θ

Solution :

tan(π/4 + θ)  =  (tanπ/4 + tanθ) / (1 - tanπ/4tanθ) :

tan(π/4 + θ)  =  (1 + tanθ) / (1 - tanθ)

tan(π/4 - θ)  =  (tanπ/4 - tanθ) / (1 + tanπ/4tanθ) :

tan(π/4 - θ)  =  (1 - tanθ) / (1 + tanθ)

Then, 

tan(π/4 + θ) - tan(π/4 - θ) :

=  [(1 + tanθ) / (1 - tanθ)] - [(1 - tanθ) / (1 + tanθ)]

=  [(1 + tanθ)² - (1 - tanθ)²] / (1 - tanθ)(1 + tanθ)

=  [(1+tan²θ+2tanθ) - (1+tan²θ-2tanθ)]/(1-tanθ)(1+tanθ)

=  4tanθ/(1 - tan²θ)

=  2[2tanθ / (1 - tan²θ)]

=  2tan2θ

Hence proved.

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