**Trigonometry Proving Problems For Class 11**

Here we are going to see some practice questions trigonometry.

**Question 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.

**Question 2 :**

Prove that (1 + tan 1^{◦})(1 + tan 2^{◦})(1 + tan 3^{◦}) ................(1 + tan 44^{◦}) is a multiple of 4.

**Solution :**

= (1 + tan 1^{◦})(1 + tan 2^{◦})(1 + tan 3^{◦}) ................(1 + tan 44^{◦})

= (1 + tan 1^{◦})(1 + tan 44^{◦})(1 + tan 2^{◦})(1 + tan 43^{◦})(1 + tan 3^{◦})(1 + tan 42^{◦})................

If A + B = 45^{◦}, then (1 + tan A)(1 + tan B) = 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.

**Question 3 :**

Prove that tan (π/4 + θ) - tan (π/4 − θ) = 2 tan 2θ.

**Solution :**

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

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

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

tan (π/4 - θ) = (1 - tan θ) / (1 + tan θ)--------(2)

(1) - (2)

tan (π/4 + θ) - tan (π/4 - θ)

= [(1 + tan θ) / (1 - tan θ)] - [(1 - tan θ) / (1 + tan θ)]

= [(1 + tan θ)^{2} -(1 - tan θ)^{2}]/(1 - tan θ)(1 + tan θ)

= [(1+tan^{2}θ+2tanθ) -(1+tan^{2}θ-2tanθ)]/(1-tan θ)(1+tan θ)

= 4tanθ/(1-tan^{2} θ)

= 2[2tanθ/(1-tan^{2} θ)]

= 2 tan 2θ

Hence proved.

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