Question

Given cot A + tan A = 5, find cot^2 A + tan^2 A

Answer 1

$\LARGE\star\boxed{\mathfrak\pink{\underline{\underline{Answer}}}}\star$

$\Large\textbf{ cot^2\:A\:+\:tan^2\:A\:=\:23 }$

$\LARGE\star\star\boxed{\mathbb\red{\underline{\underline{GIVEN}}}}\star\star$

$\Large\odot\:\:\textbf{cot\:A\:+\:tan\:A\:=\:5}$

$\LARGE\star\star\star\boxed{\mathbb\green{\underline{\underline{TO\:FIND}}}}\star\star\star$

$\Large\odot\:\:\textbf{Value\:of\: cot^2\:A\:+\:tan^2\:A }$

$\LARGE\star\star\star\star\boxed{\mathcal\red{\underline{\underline{Explanation}}}}\star\star\star\star$

$\Large\texttt{Squaring LHS and RHS}$

$\Large\texttt{Using (a+b)^2\:=\:a^2+b^2+2ab }$

$\large\Longrightarrow\textsf{ cot^2\:A\:+\:tan^2\:A\:+2\:cot\:A\:tan\:A\:=\:5^2 }$

$\Large\texttt{We know that cot A= \dfrac{1}{tan\:A} }$

$\large\Longrightarrow\textsf{ cos^2\:A\:+\:tan^2\:A\:+\:2\:\dfrac{1}{tan\:A}\times\:tan\:A\:=\:25 }$

$\large\Longrightarrow\textsf{ cos^2\:A\:+\:tan^2\:A\:+2\:=\:25 }$

$\Large\textsf{Transfering 2 from LHS to RHS}$

$\large\Longrightarrow\textsf{ cos^2\:A\:+\:tan^2\:A\:=\:23 }$

$\Large\therefore\textbf{The\:Answer\:is\:}$

$\Large\textbf{ cos^2\:A\:+\:tan^2\:A\:=\:23 }$

$\LARGE\mathcal{\underline{NOTE}}$

$\large\odot\:\:\texttt{In such types of questions}$

$\large\texttt{try to solve the question}$

$\large\texttt{by either changing every}$

$\large\texttt{term into a single term or }$

$\large\texttt{using some of the trigonometric}$

$\large\texttt{identities or relations.}$

$\Large\mathcal{\underline{Trigonometric\:Identities}}$

$\large\odot\:\:\texttt{ sin^2\:A\:+\:cos^2\:A\:=\:1 }$

$\large\odot\:\:\texttt{ cosec^2\:A\:-\:cot^2\:A\:=\:1 }$

$\large\odot\:\:\texttt{ sec^2\:A\:-\:tan^2\:A\:=\:1 }$

$\Large\mathcal{\underline{Trigonometric\:Relations}}$

$\large\texttt{sin A = \dfrac{1}{cosec\:A} }$

$\large\texttt{or cosec A = \dfrac{1}{sin\:A} }$

$\large\texttt{cos A = \dfrac{1}{sec\:A} }$

$\large\texttt{or sec A = \dfrac{1}{cos\:A} }$

$\large\texttt{tan A = \dfrac{1}{cot\:A} }$

$\large\texttt{or cot A = \dfrac{1}{tan\:A} }$

$\large\texttt{tan A = \dfrac{sin\:A}{cos\:A} }$

$\large\texttt{cot A = \dfrac{cos\:A}{sin\:A} }$

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Answer 2

Answer:

tanA + cotA = 5

(tanA +cotA)2 = tanA2 + cotA2 + 2(tanA * cotA)

25 = tanA2 + cot2A + 2

tanA *cotA = 1

tan2A + cot2A = 23

Step-by-step explanation:

hope it helps ✌️✌️