(i) Let us consider the LHS:
(cos 11° + sin 11°)/(cos 11° – sin 11°)
Let us divide the numerator and denominator by cos 11° we get,
(cos 11° + sin 11°)/(cos 11° – sin 11°) = (1 + tan 11°)/(1 – tan 11°)
= (1 + tan 11°)/(1 - 1 × tan 11°)
= (tan 45° + tan 11°)/(1 – tan 45° × tan 11°)
As we know that tan (A + B) = (tan A + tan B)/(1 – tan A tan B)
Therefore,
(tan 45° + tan 11°)/(1 – tan 45° × tan 11°) = tan (45° + 11°)
= tan 56°
= RHS
∴ LHS = RHS
Thus proved.
(ii) (cos 9° + sin 9°)/(cos 9° – sin 9°) = tan 54°
Let us consider the LHS:
(cos 9° + sin 9°)/(cos 9° – sin 9°)
Let us divide the numerator and denominator by cos 9° we get,
(cos 9° + sin 9°)/(cos 9° – sin 9°) = (1 + tan 9°)/(1 – tan 9°)
= (1 + tan 9°)/(1 – 1 × tan 9°)
= (tan 45° + tan 9°)/(1 – tan 45° × tan 9°)
As we know that tan (A + B) = (tan A + tan B)/(1 – tan A tan B)
Therefore,
(tan 45° + tan 9°)/(1 – tan 45° × tan 9°) = tan (45° + 9°)
= tan 54°
= RHS
∴ LHS = RHS
Thus proved.
(iii) Let us consider the LHS:
(cos 8° – sin 8°)/(cos 8° + sin 8°)
Let us divide the numerator and denominator by cos 8° we get,
(cos 8° – sin 8°)/(cos 8° + sin 8°) = (1 – tan 8°)/(1 + tan 8°)
= (1 – tan 8°)/(1 + 1 × tan 8°)
= (tan 45° – tan 8°)/(1 + tan 45° × tan 8°)
As we know that tan (A + B) = (tan A + tan B)/(1 – tan A tan B)
Therefore,
(tan 45° – tan 8°)/(1 + tan 45° × tan 8°) = tan (45° – 8°)
= tan 37°
= RHS
∴ LHS = RHS
Thus proved.