Question

prove all the trignometric identities​

Answer 1

{\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 \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}

Or

{\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {\left({\frac {1}{\csc \theta }}\right)}{\left({\frac {1}{\sec \theta }}\right)}}={\frac {\left({\frac {\csc \theta \sec \theta }{\csc \theta }}\right)}{\left({\frac {\csc \theta \sec \theta }{\sec \theta }}\right)}}={\frac {\sec \theta }{\csc \theta }}} \tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {\left({\frac {1}{\csc \theta }}\right)}{\left({\frac {1}{\sec \theta }}\right)}}={\frac {\left({\frac {\csc \theta \sec \theta }{\csc \theta }}\right)}{\left({\frac {\csc \theta \sec \theta }{\sec \theta }}\right)}}={\frac {\sec \theta }{\csc \theta }}

{\displaystyle \cot \theta ={\frac {\csc \theta }{\sec \theta }}} \cot \theta ={\frac {\csc \theta }{\sec \theta }}

Complementary angle identities

Two angles whose sum is π/2 radians (90 degrees) are complementary. In the diagram, the angles at vertices A and B are complementary, so we can exchange a and b, and change θ to π/2 − θ, obtaining:

{\displaystyle \sin \left(\pi /2-\theta \right)=\cos \theta } \sin \left(\pi /2-\theta \right)=\cos \theta

{\displaystyle \cos \left(\pi /2-\theta \right)=\sin \theta } \cos \left(\pi /2-\theta \right)=\sin \theta

{\displaystyle \tan \left(\pi /2-\theta \right)=\cot \theta } \tan \left(\pi /2-\theta \right)=\cot \theta

{\displaystyle \cot \left(\pi /2-\theta \right)=\tan \theta } \cot \left(\pi /2-\theta \right)=\tan \theta

{\displaystyle \sec \left(\pi /2-\theta \right)=\csc \theta } \sec \left(\pi /2-\theta \right)=\csc \theta

{\displaystyle \csc \left(\pi /2-\theta \right)=\sec \theta } \csc \left(\pi /2-\theta \right)=\sec \theta

Pythagorean identities

Edit

Identity 1:

{\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1} \sin^2(x) + \cos^2(x) = 1

The following two results follow from this and the ratio identities. To obtain the first, divide both sides of {\displaystyle \sin ^{2}(x)+\cos ^{2}(x)=1} \sin^2(x) + \cos^2(x) = 1 by {\displaystyle \cos ^{2}(x)} \cos ^{2}(x); for the second, divide by {\displaystyle \sin ^{2}(x)} \sin^2(x).

{\displaystyle \tan ^{2}(x)+1\ =\sec ^{2}(x)} \tan ^{2}(x)+1\ =\sec ^{2}(x)

{\displaystyle 1\ +\cot ^{2}(x)=\csc ^{2}(x)} 1\ +\cot ^{2}(x)=\csc ^{2}(x)

Similarly

{\displaystyle 1\ +\cot ^{2}(x)=\csc ^{2}(x)} 1\ +\cot ^{2}(x)=\csc ^{2}(x)

{\displaystyle \csc ^{2}(x)-\cot ^{2}(x)=1} {\displaystyle \csc ^{2}(x)-\cot ^{2}(x)=1}