Question

solve correct answer.. ​

Answer 1

Answer:-

$\displaystyle{\boxed{\red{\sf\:\dfrac{(\:1\:+\:\tan^2\:A\:)}{(\:1\:+\:\cot^2\:A\:)}\:=\:\tan^2\:A}}}$

Step-by-step-explanation:-

We have given a trigonometric expression.

We have to simplify it.

$\displaystyle{\sf\:\dfrac{(\:1\:+\:\tan^2\:A\:)}{(\:1\:+\:\cot^2\:A\:)}}$

$\displaystyle{\implies\sf\:\dfrac{\sec^2\:A}{cosec^2\:A}\:\:\:-\:-\:\left[\:\because\:1\:+\:\tan^2\:A\:=\:\sec^2\:A\:\&\:1\:+\:\cot^2\:A\:=\:cosec^2\:A\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{1}{\cos^2\:A}}{\dfrac{1}{\sin^2\:A}}\:\:\:-\:-\:\left[\:\sec\:A\:=\:\dfrac{1}{\cos\:A}\:\&\:cosec\:A\:=\:\dfrac{1}{\sin\:A}\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{1}{\cos^2\:A}\:\times\:\dfrac{\sin^2\:A}{1}}$

$\displaystyle{\implies\sf\:\dfrac{\sin^2\:A}{\cos^2\:A}}$

$\displaystyle{\implies\sf\:\left(\:\dfrac{\sin\:A}{\cos\:A}\:\right)^2\:\:\:-\:-\:\left[\because\:\dfrac{a^m}{b^m}\:=\:\left(\:\dfrac{a}{b}\:\right)^m\:\right]}$

$\displaystyle{\implies\sf\:\tan^2\:A\:\:\:-\:-\:\left[\:\because\:\dfrac{\sin\:A}{\cos\:A}\:=\:\tan\:A\:\right]}$

$\displaystyle{\therefore\boxed{\red{\sf\:\dfrac{(\:1\:+\:\tan^2\:A\:)}{(\:1\:+\:\cot^2\:A\:)}\:=\:\tan^2\:A}}}$

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Additional Information:-

$\displaystyle{\boxed{\begin{minipage}{4.5cm}\underline{\bf\:Trigonometric\:Identities}\\\\\sf\:1.\:\tan\:\theta\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\\\\\sf\:2.\:\sec\:\theta\:=\:\dfrac{1}{\cos\:\theta}\\\\\sf\:3.\:cosec\:\theta\:=\:\dfrac{1}{\sin\:\theta}\\\\\sf\:4.\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1\\\\\sf\:5.\:1\:+\:\tan^2\:\theta\:=\:\sec^2\:\theta\\\\\sf\:6.\:1\:+\:\cot^2\:\theta\:=\:cosec^2\:\theta\\\\\end{minipage}}$

Answer 2

Answer:

We have given a trigonometric expression.

We have to simplify it.

$\displaystyle{\sf\:\dfrac{(\:1\:+\:\tan^2\:A\:)}{(\:1\:+\:\cot^2\:A\:)}}$

$\displaystyle{\implies\sf\:\dfrac{\sec^2\:A}{cosec^2\:A}\:\:\:-\:-\:\left[\:\because\:1\:+\:\tan^2\:A\:=\:\sec^2\:A\:\&\:1\:+\:\cot^2\:A\:=\:cosec^2\:A\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{\dfrac{1}{\cos^2\:A}}{\dfrac{1}{\sin^2\:A}}\:\:\:-\:-\:\left[\:\sec\:A\:=\:\dfrac{1}{\cos\:A}\:\&\:cosec\:A\:=\:\dfrac{1}{\sin\:A}\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{1}{\cos^2\:A}\:\times\:\dfrac{\sin^2\:A}{1}}$

$\displaystyle{\implies\sf\:\dfrac{\sin^2\:A}{\cos^2\:A}}$

$\displaystyle{\implies\sf\:\left(\:\dfrac{\sin\:A}{\cos\:A}\:\right)^2\:\:\:-\:-\:\left[\because\:\dfrac{a^m}{b^m}\:=\:\left(\:\dfrac{a}{b}\:\right)^m\:\right]}$

$\displaystyle{\implies\sf\:\tan^2\:A\:\:\:-\:-\:\left[\:\because\:\dfrac{\sin\:A}{\cos\:A}\:=\:\tan\:A\:\right]}$

$\displaystyle{\therefore\boxed{\orange{\sf\:\dfrac{(\:1\:+\:\tan^2\:A\:)}{(\:1\:+\:\cot^2\:A\:)}\:=\:\tan^2\:A}}}$

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