sin0.cosec0 + sin²0.cosec²0+_______+ sin⁹0.cosec⁹0=?
Answers
Answer:
*I am using 'A' instead of Ø"
\huge\underline\mathfrak\purple{Explanation-}Explanation−
To prove :
\dfrac{SinA}{1-CosA}1−CosASinA = CosecA + CotA
Solution :
LHS : \dfrac{SinA}{1-CosA}1−CosASinA
★Rationalize it.
\implies⟹ \dfrac{SinA}{1-CosA}1−CosASinA × \dfrac{1+CosA}{1+CosA}1+CosA1+CosA
\implies⟹ \dfrac{SinA(1+CosA)}{(1-CosA)(1+CosA)}(1−CosA)(1+CosA)SinA(1+CosA)
By using,
★( a + b ) ( a - b ) = a² - b²
\implies⟹ \dfrac{SinA(1+CosA)}{1-{cos}^{2}A}1−cos2ASinA(1+CosA)
Also,
★Sin²A + Cos²A = 1
=> Sin²A = 1 - Cos²A
\implies⟹ \dfrac{SinA(1+CosA)}{{Sin}^{2}A}Sin2ASinA(1+CosA)
\implies⟹ \dfrac{\cancel{SinA}(1+CosA)}{{{Sin}^{\cancel{2}}A}}Sin2ASinA(1+CosA)
\implies⟹ \dfrac{1+CosA}{SinA}SinA1+CosA
\implies⟹ \dfrac{1}{SinA}+\dfrac{CosA}{SinA}SinA1+SinACosA
By using ratios of trigonometry :
★\dfrac{1}{SinA}SinA1 = CosecA
★ \dfrac{CosA}{SinA}SinACosA = CotA
\implies⟹ CosecA + CotA
Hence proved!
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