Question

Prove the following trigonometric identities:
tan²θcos²θ=1-cos²θ

Answer 1

Answer:

hey

Step-by-step explanation:

The main trigonometric identities between trigonometric functions are proved, using mainly the geometry of the right triangle. For greater and negative angles, see Trigonometric functions.

Elementary trigonometric identities Edit

Definitions Edit

Trigonometric functions specify the relationships between side lengths and interior angles of a right triangle. For example, the sine of angle θ is defined as being the length of the opposite side divided by the length of the hypotenuse.

The six trigonometric functions are defined for every real number, except, for some of them, for angles that differ from 0 by a multiple of the right angle (90°). Referring to the diagram at the right, the six trigonometric functions of θ are, for angles smaller than the right angle:

{\displaystyle \sin \theta ={\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}={\frac {a}{h}}} \sin \theta ={\frac {{\mathrm {opposite}}}{{\mathrm {hypotenuse}}}}={\frac {a}{h}}

{\displaystyle \cos \theta ={\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}={\frac {b}{h}}} \cos \theta ={\frac {{\mathrm {adjacent}}}{{\mathrm {hypotenuse}}}}={\frac {b}{h}}

{\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {a}{b}}} \tan \theta ={\frac {{\mathrm {opposite}}}{{\mathrm {adjacent}}}}={\frac {a}{b}}

{\displaystyle \cot \theta ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}={\frac {b}{a}}} \cot \theta ={\frac {{\mathrm {adjacent}}}{{\mathrm {opposite}}}}={\frac {b}{a}}

{\displaystyle \sec \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}={\frac {h}{b}}} \sec \theta ={\frac {{\mathrm {hypotenuse}}}{{\mathrm {adjacent}}}}={\frac {h}{b}}

{\displaystyle \csc \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}={\frac {h}{a}}} \csc \theta ={\frac {{\mathrm {hypotenuse}}}{{\mathrm {opposite}}}}={\frac {h}{a}}

Ratio identities Edit

In the case of angles smaller than a right angle, the following identities are direct consequences of above definitions through the division identity

{\displaystyle {\frac {a}{b}}={\frac {\left({\frac {a}{h}}\right)}{\left({\frac {b}{h}}\right)}}.} {\displaystyle {\frac {a}{b}}={\frac {\left({\frac {a}{h}}\right)}{\left({\frac {b}{h}}\right)}}.}

They remain valid for angles greater than 90° and for negative angles.

{\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {\left({\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}\right)}{\left({\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}\right)}}={\frac {\sin \theta }{\cos \theta }}} \tan \theta ={\frac {{\mathrm {opposite}}}{{\mathrm {adjacent}}}}={\frac {\left({\frac {{\mathrm {opposite}}}{{\mathrm {hypotenuse}}}}\right)}{\left({\frac {{\mathrm {adjacent}}}{{\mathrm {hypotenuse}}}}\right)}}={\frac {\sin \theta }{\cos \theta }}

{\displaystyle \cot \theta ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}={\frac {\left({\frac {\mathrm {adjacent} }{\mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {opposite} }{\mathrm {adjacent} }}\right)}}={\frac {1}{\tan \theta }}={\frac {\cos \theta }{\sin \theta }}} \cot \theta =\frac{\mathrm{adjacent}}{\mathrm{opposite}}

= \frac { \left( \frac{\mathrm{adjacent}}{\mathrm{adjacent}} \right) } { \left( \frac {\mathrm{opposite}}{\mathrm{adjacent}} \right) }

= \frac {1}{\tan \theta} = \frac {\cos \theta}{\sin \theta}

{\displaystyle \sec \theta ={\frac {1}{\cos \theta }}={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}} \sec \theta ={\frac {1}{\cos \theta }}={\frac {{\mathrm {hypotenuse}}}{{\mathrm {adjacent}}}}

{\displaystyle \csc \theta ={\frac {1}{\sin \theta }}={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}} \csc \theta ={\frac {1}{\sin \theta }}={\frac {{\mathrm {hypotenuse}}}{{\mathrm {opposite}}}}

{\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {\left({\frac {\mathrm {opposite} \times \mathrm {hypotenuse} }{\mathrm {opposite} \times \mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {adjacent} \times \mathrm {hypotenuse} }{\mathrm {opposite} \times \mathrm {adjacent} }}\right)}}={\frac {\left({\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}\right)}}={\frac {\sec \theta }{\csc \theta }}} \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}}

= \frac{\left(\frac{\mathrm{opposite} \times \

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Answer 2

Proving:

Step-by-step explanation:

$\ Given \ value: \\\\\tan^2 \theta \cos^2\theta \ =\ 1 - \cos^2 \theta \\\\\ Solution: \\\\\tan^2 \theta \cos^2\theta \ =\ 1 - \cos^2 \theta \\\\\tan^2 \theta \ = \frac{\sin^2 \theta}{\cos^2 \theta}\\\\\ Solve \ L.H.S \ part: \\\\\rightarrow \frac{\sin^2 \theta}{\cos^2 \theta}\cdot \cos^2\theta \\\\\rightarrow \sin^2 \theta \\\\\ formula: \\\\\therefore \ \ \sin^2\theta \ + \ \cos^2\theta \ = 1\\\\\sin^2\theta \ = \ 1\ - \cos^2\theta \\\\ \rightarrow \ 1\ - \cos^2\theta$

$\ L.H.S \ = R.H.S$

Learn more:

• Proving: https://brainly.in/question/8112701