(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.