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

$Evaluate; frac{tan ;50^0+ sec;50^0}{cot;40^0+cosec;40^0}+ cos;40^0cosec;50^0$
A. 1	B. 2	C. 0	D. None of these
Right Option : B
View Explanation

Question : 2

$cos;81^0-sin;9^0$ = __________________
A. 0	B. 1	C. 2	D. $large frac{1}{4}$
Right Option : A
View Explanation

Question : 3

Find the value of $frac{tan;40^0}{cot;50^0}+ tan;5^0tan;10^0tan;30^0tan;80^0tan;85^0$
A. $frac{1}{sqrt3}$	B. $2- frac{1}{sqrt3}$	C. $1+ frac{1}{sqrt3}$	D. None of these
Right Option : C
View Explanation

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Trigonometric ratios of Complementary Angles

Trigonometric ratios of Complementary Angles

Students / Parents Reviews [20]

Barkha Arora

Aman Kumar Shrivastava

Manya

Hiya Gupta

Shreya Shrivastava

Yogansh Nyasi

Cheshta

Shivam Rana

Prisha Gupta

Aman Kumar Shrivastava

Manish Kumar

Manish Kumar

Om Umang

Barkha Arora

Anika Saxena

Yatharthi Sharma

Ayan Ghosh

Aman Kumar Shrivastava

Sharandeep Singh

Upmanyu Sharma