Question

Evaluate the following :
(v)$\frac{tan35^{0}}{cot55^{0}}+\frac{cot78^{0}}{tan12^{0}}-1$
(vi)$\frac{sec70^{0}}{cosec20^{0}}+\frac{sin59^{0}}{cos31^{0}}$
(vii)cosec 31° − sec 59°
(viii)(sin 72° + cos 18°) (sin 72° − cos 18°)

Answer 1

(v) SOLUTION :  

Given :  

tan 35° /cot 55° + cot 78°/tan 12° −1

= tan (90° - 55°) / cot 55° + cot (90° - 12°) /tan 12°  - 1

= cot 55° / cot 55° + tan 12° / tan 12° - 1

[tan (90° – θ) = cot θ and cot(90° – θ) = tanθ]

= 1 + 1 -1

= 2 -1

tan 35° /cot 55° + cot 78°/tan 12° −1 = 1

Hence, the  value of tan 35° /cot 55° + cot 78°/tan 12° − 1 is 1.

 

(vi) SOLUTION :  

Given :  sec 70° / cosec 20° + sin 59° / cos31°

sec 70° / cosec 20° + sin 59° / cos 31° = (sec(90°−20°)/ cosec 20°) – (sin(90°−31°)/cos 31°)

= cosec20° /cosec20°+ cos31°/cos31°  

[sec(90° − θ) = cosec θ and sin (90 - θ) = Cos θ]

=1 + 1  

= 2

Hence, the value of sec 70° / cosec 20° + sin 59° / cos31° is 2.

 

(vii) SOLUTION :  

Given :  cosec 31°− sec 59°

cosec 31°− sec 59° = cosec(90°−59°)− sec59°  

=sec 59°− sec 59°

[cosec(90° − θ)=sec θ]

cosec 31°− sec 59° = 0

Hence, the value of cosec 31°− sec 59° is 0.

 

(viii) SOLUTION :  

Given : (sin 72° + cos 18°) (sin 72° − cos 18°)  

(sin 72° + cos 18°) (sin 72° − cos 18°) = (sin 72°)² – (cos 18°)²

[(a+b) (a -b) = a² - b²]

= sin(90°−18°)² −(cos 18°)²

= (cos18°)²– (cos18°)²

[sin (90 - θ) = Cos θ]

=cos²18°− cos²18° = 0

(sin 72° + cos 18°) (sin 72° − cos 18°) = 0

Hence, the value of (sin 72° + cos 18°) (sin 72° − cos 18°) is 0.

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Answer 2

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