Question

Prove that :
(i)tan 20° tan 35° tan 45° tan 55° tan 70° = 1
(ii)sin 48° sec 42° + cos 48° cosec 42° = 2
(iii)$\frac{sin70^{0}}{cos20^{0}}+\frac{cosec20^{0}}{sec70^{0}}-2cos70^{0}cosec20^{0}=0^{0}$
(iv)$\frac{cos80^{0}}{sin10^{0}}+cos59^{0}cosec31^{0}=2$

Answer 1

(i) SOLUTION :  

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 2

hope this helps. ...