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Trigonometric ratios of Complementary Angles

Introduction to Trigonometry II - Concepts

Trigonometric ratios of Complementary Angles

Trigonometric Identity Involving Sine And Cosine

Trigonometric Identity Involving Secant And Tangent

Trigonometric Identity Involving Cosecant And Cotangent

Trigonometric Identities

Class - RPF Constable Subjects

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

$cos;48^0-sin;42^0=$ ___________________
A. 0	B. 1	C. 2	D. $frac{1}{4}$
Right Option : A
View Explanation

Question : 2

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

Question : 3

$cos^2;57^0-sin^2;33^0=$ ___________________
A. 0	B. 1	C. 2	D. $frac{1}{4}$
Right Option : A
View Explanation

Chapters

Exponents and Power II

Introduction to Trigonometry II

Introduction to Trigonometry I

Ratio and Proportion

Quadrilaterals II

Quadrilaterals I

Linear Equations in One Variable II

Linear Equations in One Variable I

Cube and Cube Roots

Square and Square Root II

Square and Square Root I

Rational Numbers II

Rational Numbers I

Exponents and Power I

Direct and Inverse Variation II

Direct and Inverse Variation I

Algebraic Identities

Algebraic Expressions

Comparing Quantities III

Comparing Quantities II

Comparing Quantities I

Playing With Numbers

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