Question

tanA + cot A = sec a . cosec A
prove it :

Answer 1

Solution :

Tan A + Cot A

> Sin A/cos A + cos A/sin A

Taking LCM

> [ ( sin A )( sin A ) + ( cos A)( cos A) ]/ sin A cos A

> [ sin ² A + cos ² A ]/ sin A cos A .

We know that sin ²A + cos ² A = 1

> 1/sin A cos A

> [ 1/sin A ] × [ 1/cos A ]

> [ cosec A ] × [ sec A]

> sec A cosec A .

Hence Proved !

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Additional Information :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$

Answer 2

tanA + cot A = sec a . cosec A

tan A = sin A/cos A and cot A = cos A/sin A

$\dashrightarrow\rm \dfrac{sin \ A}{cos \ A} + \dfrac{cos \ A}{sin \ A} = \sec A \cosec A$

$\dashrightarrow\rm \dfrac{(sin \ A)(sin \ A)+(cos \ A)(cos \ A)}{cos \ A \sin \ A} = \sec A \cosec A$

$\dashrightarrow\rm \dfrac{sin^2\ A+cos^2 \ A}{cos \ A \ sin \ A} = \sec A \cosec A$

sin² A + cos² A = 1

$\dashrightarrow\rm \dfrac{1}{cos \ A \sin A} = \sec A \cosec A$

$\dashrightarrow\rm \dfrac{1}{cos \ A} \times \dfrac{1}{ sin \ A} = \sec A \cosec A$

$\bf sec \ A = \dfrac{1}{cos \ A} \\ \\ \bf cosec \ A = \dfrac{1}{sin \ A}$

$\dashrightarrow\rm \sec A \cosec A = \sec A \cosec A$

Hence, proved.