Question

secA divided by tanA + cotA is equal to?​

Answer 1

Answer:

sinA

Step-by-step explanation:

secA/(tanA+cotA)

=(1/cosA) /{(sinA/cosA) +(cosA/sinA) }

=(1/cosA) /{(sin^2A+cos^2A)/(sinAcosA)}

=(1×(sinAcosA))/(1×cosA)

(since sin^2A+cos^2A=1)

=(sinAcosA) /cosA

=sinA

Answer 2

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• $\sf \dfrac{ \sec A }{\tan A + \cot A}$

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• $\sf Value \: of \: \dfrac{ \sec A }{\tan A + \cot A}$

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

$\underline{\:\textsf{Given that :}}$

• $\sf \dfrac{ \sec A }{\tan A + \cot A}$

$\underline{\:\textsf{Then:}}$

$\sf \dashrightarrow \dfrac{ \sec A }{\tan A + \cot A}$

$\sf \dashrightarrow \dfrac{ \sec A }{ \dfrac{ \sin A}{\cos A} + \dfrac{ \cos A}{\sin A}}$

$\qquad\qquad\qquad$

$\sf \dashrightarrow \dfrac{ \sec A }{ \dfrac{ {\sin}^{2} A + {\cos}^{2} A }{\cos A \sin A } }$

$\sf \dashrightarrow \dfrac{ \sec A }{ \dfrac{ {\sin}^{2} A + {\cos}^{2} A }{\cos A \sin A } }$

$\sf \dashrightarrow \dfrac{ \sec A }{ \dfrac{1}{\cos A \sin A } }$

$\sf \dashrightarrow \sec A \times \dfrac{\cos A \sin A}{1}$

$\sf \dashrightarrow \sec A \times \cos A \sin A$

$\sf \dashrightarrow \dfrac{1}{\cos A} \times \cos A \sin A$

$\sf \dashrightarrow \dfrac{1}{ \cancel{\cos A}} \times \cancel{ \cos A} \sin A$

$\sf \dashrightarrow 1 \times \sin A$

$\sf \dashrightarrow \sin A$

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\therefore{ \sf \dfrac{ \sec A }{\tan A + \cot A} = \sin A }$

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ㅤ$\;\;\underline{\textbf{\textsf{ Need to know :-}}}$

• $\sf{ tanA= \dfrac{sinA}{cosA} }$

• $\sf{ cotA= \dfrac{cosA}{sinA} }$

• $\sf{ secA= \dfrac{1}{cosA} }$

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