can anyone help me in this question
Answers
Answer:
a = 2² = -4
b = 2 / 5 ²= 4 / 25
c = 1 / 6 ⁴ = 1 / 1296
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Answer:
\large \bf \clubs \: To \: Find :-♣ ToFind:−
\small\bf\dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA} =sinA - {cosA}
1+cotA
sinA
−
1+tanA
cosA
=sinA−cosA
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\large \bf \clubs \: Formulae \: Used :-♣ FormulaeUsed:−
-----------------------
\begin{gathered}\bf\maltese\:\:\: tanA=\dfrac{sinA}{cosA} \\ \\ \bf\maltese\:\:\:cotA=\dfrac{cosA}{sinA} \\ \\\bf\maltese\:\:\:a^{2} -b^{2} =(a-b) (a+b)\end{gathered}
✠tanA=
cosA
sinA
✠cotA=
sinA
cosA
✠a
2
−b
2
=(a−b)(a+b)
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\large \bf \clubs \: Proof :-♣ Proof:−
\LARGE\bf{LHS} : -LHS:−
\begin{gathered}\sf\dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA} \\ \\ \sf=\dfrac{sinA}{1 + \dfrac{cosA}{sinA} } - \dfrac{cosA}{1 + \dfrac{sinA}{cosA} }\\ \\\sf=\dfrac{sin^2A}{sinA + cosA} - \dfrac{cos^{2}A }{cosA + sinA} \\ \\ \sf=\dfrac{sin^2A-cos^2A}{sinA + cosA} \\ \\ \sf = \frac{ \cancel{(sinA + cosA)}(sinA - cosA)}{ \cancel{sinA + cosA}} \\ \\ \bf= sinA + cosA\end{gathered}
1+cotA
sinA
−
1+tanA
cosA
=
1+
sinA
cosA
sinA
−
1+
cosA
sinA
cosA
=
sinA+cosA
sin
2
A
−
cosA+sinA
cos
2
A
=
sinA+cosA
sin
2
A−cos
2
A
=
sinA+cosA
(sinA+cosA)
(sinA−cosA)
=sinA+cosA
\LARGE=\bf{RHS}=RHS
\Large \purple{ \bf\bigstar \: Hence \: Proved \: \bigstar}★HenceProved★
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\large \bf \clubs \: Additional \: Info :-♣ AdditionalInfo:−
\begin{gathered}\qquad \pink{\bigstar \: \bf Fundamental \: Trigonometric\:\bigstar} \\ \qquad\bf\pink{\bigstar\:Identities\:\bigstar}\end{gathered}
★FundamentalTrigonometric★
★Identities★
\begin{gathered}\large\qquad \boxed{\boxed{\begin{array}{cc} \maltese \: \sf \: { \sin }^{2} \theta + { \cos}^{2} \theta = 1 \\ \\ \maltese\sf \: \: { \sec}^{2} \theta = 1 + { \tan}^{2} \theta \\ \\ \maltese \: \sf \: { \cosec}^{2} \theta = 1 + { \cot}^{2} \theta \end{array}}}\end{gathered}
✠sin
2
θ+cos
2
θ=1
✠sec
2
θ=1+tan
2
θ
✠cosec
2
θ=1+cot
2
θ
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