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Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum is 90°.

 

Thus, θ and 90° - θ are complementary angles.
In the above triangle ABC, we get,                                       
   sin
θ = and cos (90° - θ) =
   cos
θ = and sin (90° - θ) =
   tan
θ = and cot (90° - θ) =
cosec
θ = and sec (90° - θ) = 
 sec
θ = and cosec (90° - θ) = 
   cot
θ = and tan (90° - θ) =
 
Therefore, we can say that,
       sin (90° - 
θ) = cos θ
       cos (90° -  q) = sin θ
       tan (90° -  q) = cot θ
    cosec (90° -  q) = sec θ
       sec (90° - θ) = cosec θ
        cot(90° -  q) = tan θ

 

Example

Express the following in terms of the angles between 0° and 45°.
(i)  sin 81° + cos 82°
(ii) sin 75° + tan 70°
(iii)cos 80° + sec 75°
(iv)sec 65° + cot 70°
(v)cosec 80° + cosec 70°

Solution

(i) sin 81° + cos 82°
               = sin (90° - 9°) + cos (90° - 8°)    [
Q cos θ = sin (90° - θ) and sin θ                =  cos (90° - θ)] 
               = cos 9° + sin 8°
 
(ii) sin 75° + tan 70°
              = sin (90° - 15°) + tan (90° - 20°)    [
Q cos θ = sin (90° - θ) and cot θ               = tan (90° - θ)] 
              = cos 15° + cot 20°
 
(iii) cos 80° + sec 75°
              = cos (90° - 10°) + sec (90° - 15°)    [
Q sin θ = cos (90° - θ) and cosec θ               = sec (90° - θ)] 
              = sin 10° + cosec 15°
 
(iv) sec 65° + cot 70°
            = sec (90° - 25°) + cot (90° - 20°)         [
Q cosec θ = sec (90° - θ) and tan θ
            = cot (90° - θ)] 
            = cosec 25° + tan 20°
 
(v) cosec 80° + cosec 70°
             = cosec (90° - 10°) + cosec (90° - 20°)    [
Q cosec θ = sec (90° - θ)]
             = sec 10° + sec 20°

 





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