Prove that:
(i)cos 80°/sin 10° + cos 59°. cosec 31°= 2
(ii) cos 90°-Θ/ 1 + sin(90° - Θ) + 1 + sin(90° - Θ)/ cos(90° - Θ) = 2cosecΘ
(iii)sec(90° - Θ) cosecΘ - tan(90° - Θ) cot Θ + cos² 25° + cos² 65°/ 3 tan 27° tan 63° - 2/3
Answers
Step-by-step explanation:
Given : tan 20° tan 35° tan 45° tan 55° tan 70° = 1
L.H.S = tan 20° tan 35° tan 45° tan 55° tan 70°
= tan(90°–70°) tan(90°–55°) tan 45°tan 55°tan 70°
= cot70°cot55°tan45°tan55°tan70°
[tan θ = cot (90 - θ) ]
= (tan70°cot70°)(tan55°cot55°)tan45°
=1 x 1 x 1
[tan θ × cot θ = 1, tan 45° = 1 ]
= 1
= RHS
tan 20° tan 35° tan 45° tan 55° tan 70° = 1
Hence proved
(ii) SOLUTION :
Given : sin 48° sec 42° + cos 48° cosec 42° = 2
LHS = sin 48° sec 42° + cos 48° cosec 42°
= sin 48° × sec(90°− 48°)cos48° × cosec(90°− 48°)
= sin48° × cosec48° + cos48° × sec48°
[sec (90 - θ) = cosθ, cosec (90 - θ) = sec θ ]
= 1 + 1
[sin θ × cosec θ = 1 , cos θ × sec θ = 1]
= 2
RHS
sin 48° sec 42° + cos 48° cosec 42° = 2
Hence proved
(iii) SOLUTION :
Given : sin70°/cos20° + cosec20°/sec70°− 2cos70° cosec20° = 0
LHS = sin70°/cos20° + cosec 20°/sec 70°− 2cos70° cosec 20°
= sin(90°−20°)/cos 20°+ cos(90°−20°)/sin 20°– 2cos70°.cosec(90°−70°)
= cos20°/cos20°+ sin20°/sin20° −2 × cos 70° × sec 70°
[sin (90 - θ) = cos θ, cos θ= sin (90 - θ) ,cosec (90 - θ) = sec θ ,cosθ × sec θ = 1]
= 1 + 1-2
= 2 - 2
= 0
= RHS
sin70°/cos20° + cosec20°/sec70°− 2cos70° cosec20° = 0
Hence proved
(iv) SOLUTION :
Given : cos 80°/sin 10° + cos 59° cosec 31°= 2
LHS = cos 80°/sin 10° + cos 59° cosec31°
= cos(90°−10°)/sin 10°+ cos 59° cosec(90°−59°)
= sin10°/sin10° + cos59° sec59°
[cos (90 - θ) = sin θ, cosec (90 - θ) = sec θ ]
= 1+1
[cos θ . sec θ = 1]
= 2
= RHS
cos 80°/sin 10° + cos 59° .cosec 31°= 2
Hence proved.
Answer:
Step-by-step explanation:
(i) cos 80°/sin 10° + cos 59° . cosec 31° = 2
LHS
= cos(90°-80°)/sin 10° + cos (90°-59°) .cosec 31°
= sin 10°/sin 10° + sin 31° . cosec 31°
= 1 + sin 31° . 1/sin 31°
= 1+1
= 2
HENCE PROVED