In the isosceles triangle A B C , B D is the altitude and ∠ A B C = 120 ∘ . What is the value of cos C ?

∵ A B C is an isosceles triangle and ∠ A B C = 120 ∘ ∴ ∠ A = ∠ C = 30 ∘ Now in right angled triangle B C D Now cos C = cos 30 ∘ = 3 2

Prove that tan A 1 - cot A + cot A 1 - tan A = 1 + tan A + cot A = 1 + sec A cosec A .

Consider tan A 1 - cot A + cot A 1 - tan A = 1 + tan A + cot A L.H.S. = tan A 1 - cot A + cot A 1 - tan A = tan A 1 - 1 tan A + 1 tan A 1 - tan A = tan 2 A tan A - 1 + 1 tan A 1 - tan A = - tan 2 A 1 - tan A + 1 tan A 1 - tan A = - tan 3 A + 1 tan A 1 - tan A = 1 - tan 3 A tan A 1 - tan A = 1 - tan A 1 + tan A + tan 2 A tan A 1 - tan A = 1 + tan A + tan 2 A tan A = 1 tan A + tan A tan A + tan 2 A tan A = 1 + tan A + cot A Now, consider tan A 1 - cot A + cot A 1 - tan A = 1 + sec A · Cosec A L.H.S. = tan A 1 - cot A + cot A 1 - tan A = sin A cos A 1 - cos A sin A + cos A sin A 1 - sin A cos A = sin A cos A sin A - cos A sin A + cos A sin A cos A - sin A cos A = sin 2 A cos A sin A - cos A + cos 2 A sin A cos A - sin A = sin 2 A cos A sin A - cos A - cos 2 A sin A sin A - cos A = sin 3 A cos A sin A sin A - cos A - cos 3 A sin A cos A sin A - cos A = sin 3 A - cos 3 A cos A sin A sin A - cos A = sin A - cos A sin 2 A + sin A cos A + cos 2 A cos A sin A sin A - cos A = sin 2 A + sin A cos A + cos 2 A cos A sin A = 1 + sin A cos A cos A sin A = sec A cosec A + 1

If sin θ + cos θ = 3 , then prove that tan θ + c o t θ = 1 . cnsnbcnsbcjksbcjb

Given that, sin θ + cos θ = 3 Squaring on both sides, sin θ + cos θ 2 = 3 2 sin 2 θ + cos 2 θ + 2 sin θ cos θ = 3 1 + 2 sin θ cos θ = 3 [using sin 2 A + cos 2 A = 1 ] 2 sin θ cos θ = 2 sin θ cos θ = 1 ........................(i) Now consider, tan θ + co t θ = sin θ cos θ + cos θ sin θ = sin 2 θ + cos 2 θ sin θ cos θ = 1 sin θ cos θ = 1 1 [from (i)] = 1 ∴ tan θ + c o t θ = 1 Hence, proved.