Trigonometric ratios of Complementary Angles


 
 
Concept Explanation
 

Trigonometric ratios of Complementary Angles

Trigonometric ratios of Complementary Angles:

Recall that two angles are said to be complementary if their sum equals 90^{circ}. In Delta ABC, right - angled at B, do you see any pair of complementary angles?

Since    angle A+angle C=90^{circ},they form such a pair, we have:

  sin A=frac{BC}{AC}                  cos A=frac{AB}{AC}                 tan A=frac{BC}{AB}

cosec A=frac{AC}{BC}          sec A=frac{AC}{AB}                  cot A=frac{AB}{BC}             (1)

Now let us write the trigonometric ratios for  angle C=90^{circ}-angle A.

For convenience, we shall write 90^{circ}-A instead of 90^{circ}-angle A.

What would be the side opposite and the side adjacent to the angle 90^{circ}- A?

You will find that AB is the side opposite and BC is the side adjacent to the angle 90^{circ}- A. Therefore,

sin(90^{circ}-A)=frac{AB}{AC}        cos(90^{circ}-A)=frac{BC}{AC}         tan(90^{circ}-A)=frac{AB}{BC}

cosec(90^{circ}-A)=frac{AC}{AB}      sec(90^{circ}-A)=frac{AC}{BC}      cot(90^{circ}-A)=frac{BC}{AB}              ....... (2)

Now, compare the ratios in (1) and (2) , observe that :

sin(90^{circ}-A)=frac{AB}{AC}=cos A;;and;;cos(90^{circ}-A)=frac{BC}{AC}=sin A

Also,   tan(90^{circ}-A)=frac{AB}{BC}=cot ;A;;and;;cot;(90^{circ}-A)=frac{BC}{AB}=tan ;A

sec;(90^{circ}-A)=frac{AB}{BC}=cosec ;A;;and;;cosec;(90^{circ}-A)=frac{AC}{AB}=sec; A

So,   sin;(90^{circ}-A)=cos; A,                    cos;(90^{circ}-A)=sin; A,

       tan;(90^{circ}-A)=cot ;A,                    cot;(90^{circ}-A)=tan; A

      sec;(90^{circ}-A)=cosec ;A              cosec;(90^{circ}-A)=sec ;A

for all values of angle A lying between 0^{circ}  and  90^{circ}.  check whether this holds for A=0^{circ}  or  A=90^{circ}.

Example 1:  Evaluate frac{tan; 65^{circ}}{cot; 25^{circ}}

Solution:  We know:       cot ;A = tan; (90^{circ}-A)

So,                             cot; 25^{circ} = tan ;(90^{circ}-25^{circ})=tan;65^{circ}

i.e,                             frac{tan;65^{circ}}{cot;25^{circ}}-=frac{tan;65^{circ}}{tan;65^{circ}}=1

Sample Questions
(More Questions for each concept available in Login)
Question : 1

Find the value of

frac{sin;65^0}{cos;25^0}+ sin;5^0 cos;10^0 tan;45^0cosec;80^0sec;85^0

Right Option : B
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Explanation
Question : 2

Evaluate ;cosec ;(65^0+theta)- sec ;(25^0-theta)

Right Option : C
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Explanation
Question : 3

Evaluate: cos ;48^0-sin; 42^0

Right Option : A
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Explanation
 
 
 


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