Prove:sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan

1.	Prove:sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 70° tan 80° tan 89° = 1
Answer» Taking the L.H.S, = sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = sin (50° + θ) – sin (90° – (40° – θ)) + tan (90 – 89)° tan (90 – 80)° tan (90 – 70)° tan 70° tan 80° tan 89° [∵ sin (90 – θ) = cos θ] = sin (50° + θ) – sin (50° + θ) + cot 89° cot 80° cot 70° tan 70° tan 80° tan 89° [∵ tan (90° – θ) = cot θ] = 0 + (cot 89° x tan 89°) (cot 80° x tan 80°) (cot 70° x tan 70°) = 0 + 1 x 1 x 1 [∵ tan θ x cot θ = 1] = 1 = R.H.S Hence Proved

Prove:sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 70° tan 80° tan 89° = 1

Answer»

Taking the L.H.S,

= sin (50° + θ) – cos (40° – θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89°

= sin (50° + θ) – sin (90° – (40° – θ)) + tan (90 – 89)° tan (90 – 80)° tan (90 – 70)° tan 70° tan 80° tan 89° [∵ sin (90 – θ) = cos θ]

= sin (50° + θ) – sin (50° + θ) + cot 89° cot 80° cot 70° tan 70° tan 80° tan 89° [∵ tan (90° – θ) = cot θ]

= 0 + (cot 89° x tan 89°) (cot 80° x tan 80°) (cot 70° x tan 70°)

= 0 + 1 x 1 x 1 [∵ tan θ x cot θ = 1]

= 1

= R.H.S

Hence Proved