Question

Evaluate the following :
(i)$(\frac{sin49^{0} }{cos41^{0} } )^{2}+(\frac{cos41^{0} }{sin49^{0} })^{2}$2
(ii)cos 48° − sin 42°
(iii)$\frac{cot40^{0} }{tan50^{2} }-\frac{1}{2} (\frac{cos35^{0} }{sin55^{0}})$
(iv)$(\frac{sin27^{0} }{cos63^{0} } )^{2}-(\frac{cos63^{0} }{sin27^{0} })^{2}$

Answer 1

SOLUTION :  

Given :  (sin 49° / cos 41°)² + (cos 41°/sin 49°)²

= (sin(90°− 41°)/cos 41°)² +(cos(90° −49°)/sin49°)²

[sin(90∘ – θ) = cos θ and cos(90° – θ) = sin θ]

= (cos 41° / cos 41°)² + (sin 49° /sin49°)²

= 1² +1² = 1 + 1 = 2

(sin 49° / cos 41°)² + (cos 41°/sin 49°)² = 2

Hence, the value of (sin 49° / cos 41°)² + (cos 41°/sin 49°)² is 2.  

 

(ii) SOLUTION :

Given : cos 48° – sin 42°

cos 48° – sin 42° = cos(90°− 42°)− sin 42°

[cos(90° − θ) = sinθ]

= sin 42° – sin 42°  = 0

cos 48° – sin 42° = 0

Hence,the value of cos 48° – sin 42° is 0.

 

(iii) SOLUTION :  

Given :  cot 40°/tan 50°−1/2(cos 35°/sin 55°)

cot 40°/tan 50°−1/2(cos 35°/sin 55°) = cot(90° −50°) tan 50° −1/2(cos(90°−55°)/sin 55°)

= tan 50°/ tan 50° −1/2(sin 55°/sin 55°)

[cot(90° − θ ) =  tan θ and cos(90° −θ)= sin θ]

= 1−1/2  

= (2 - 1)/2

= 1/2

cot 40°/tan 50°−1/2(cos 35°/sin 55°)  = 1/2

Hence, the value of cot 40°/tan 50°−1/2(cos 35°/sin 55°) is 1/2.

 

(iv) SOLUTION :  

Given : (sin 27°/cos 63°)² – (cos 63°/sin 27°)²

[sin(90∘ – θ) = cos θ and cos(90° – θ) = sin θ]

(sin 27°/cos 63°)² – (cos 63°/sin 27°)² =  (sin(90° −63° )/cos63°)² – (cos(90° −27°)/sin 27∘)²

=(cos63°/cos63°)² – (sin27°/sin27°)²  

= 1 - 1  = 0

(sin 27°/cos 63°)² – (cos 63°/sin 27°)² = 0

Hence, the value of (sin 27°/cos 63°)² – (cos 63°/sin 27°)²  is 0.

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