Appropriate Question :- Solve
[tex]\sf \: 5 \: sin50 \degree \: sec40\degree - 3 \: cos59\degree \: cosec31\degree \\ \\ [/tex]
Answer:
[tex]\qquad\boxed{ \sf{ \:\sf \: 5 \: sin50 \degree \: sec40\degree - 3 \: cos59\degree \: cosec31\degree = 2 \: }} \\ \\ [/tex]
Step-by-step explanation:
Given trigonometric expression is
[tex]\sf \: = \: 5 \: sin50 \degree \: sec(90\degree - 50\degree) - 3 \: cos59\degree \: cosec(90\degree - 59\degree) \\ \\ [/tex]
[tex]\sf \: = \: 5 \: sin50 \degree \: cosec50\degree - 3 \: cos59\degree \: sec59\degree \\ \\ [/tex]
[tex]\sf \: = \: 5 \: sin50 \degree \times \dfrac{1}{sin50\degree} - 3 \: cos59\degree \times \dfrac{1}{cos59\degree} \\ \\ [/tex]
[tex]\sf \: = \: 5 - 3 \\ \\ [/tex]
[tex]\sf \: = \: 2 \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: 5 \: sin50 \degree \: sec40\degree - 3 \: cos59\degree \: cosec31\degree = 2 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
[tex]\sf \: sec(90\degree - x) = cosecx \\ \\ [/tex]
[tex]\sf \: cosec(90\degree - x) = secx \\ \\ [/tex]
[tex]\sf \: cosecx = \dfrac{1}{sinx} \\ \\ [/tex]
[tex]\sf \: secx = \dfrac{1}{cosx} \\ \\ [/tex]
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \\ [/tex]
Correct Question :
[tex] \sf5 \sin50 \degree× \sec40 \degree-3 \cos59 \degree × \cosec 31 \degree[/tex]
Solution :
[tex] \qquad \sf = 5 \sin50 \degree× \sec40 \degree-3 \cos59 \degree × \cosec 31 \degree[/tex]
[tex] \qquad \sf = 5 \sin50 \degree× \sec(90 - 50) \degree-3 \cos59 \degree × \cosec (90 - 59) \degree[/tex]
[tex] \qquad \sf = 5 \sin50 \degree× \cosec 50 \degree-3 \cos59 \degree × \sec59\degree[/tex]
[tex] \displaystyle\qquad \sf = 5 \: \cancel{ \sin50 }\degree× \frac{1}{ \cancel{\sin50}\degree} -3 \: \cancel{\cos59 }\degree × \frac{1}{ \cancel{\cos59}\degree} [/tex]
[tex] \displaystyle\qquad \sf = \: 5 - 3[/tex]
[tex] \red{ \displaystyle\qquad \bf = \: 2}[/tex]
[tex] \red{ \sf \therefore \sf5 \sin50 \degree× \sec40 \degree-3 \cos59 \degree × \cosec 31 \degree = 2}[/tex]
[tex] \: [/tex]
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Answers & Comments
Appropriate Question :- Solve
[tex]\sf \: 5 \: sin50 \degree \: sec40\degree - 3 \: cos59\degree \: cosec31\degree \\ \\ [/tex]
Answer:
[tex]\qquad\boxed{ \sf{ \:\sf \: 5 \: sin50 \degree \: sec40\degree - 3 \: cos59\degree \: cosec31\degree = 2 \: }} \\ \\ [/tex]
Step-by-step explanation:
Given trigonometric expression is
[tex]\sf \: 5 \: sin50 \degree \: sec40\degree - 3 \: cos59\degree \: cosec31\degree \\ \\ [/tex]
[tex]\sf \: = \: 5 \: sin50 \degree \: sec(90\degree - 50\degree) - 3 \: cos59\degree \: cosec(90\degree - 59\degree) \\ \\ [/tex]
[tex]\sf \: = \: 5 \: sin50 \degree \: cosec50\degree - 3 \: cos59\degree \: sec59\degree \\ \\ [/tex]
[tex]\sf \: = \: 5 \: sin50 \degree \times \dfrac{1}{sin50\degree} - 3 \: cos59\degree \times \dfrac{1}{cos59\degree} \\ \\ [/tex]
[tex]\sf \: = \: 5 - 3 \\ \\ [/tex]
[tex]\sf \: = \: 2 \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: 5 \: sin50 \degree \: sec40\degree - 3 \: cos59\degree \: cosec31\degree = 2 \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
[tex]\sf \: sec(90\degree - x) = cosecx \\ \\ [/tex]
[tex]\sf \: cosec(90\degree - x) = secx \\ \\ [/tex]
[tex]\sf \: cosecx = \dfrac{1}{sinx} \\ \\ [/tex]
[tex]\sf \: secx = \dfrac{1}{cosx} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \\ [/tex]
Correct Question :
[tex] \sf5 \sin50 \degree× \sec40 \degree-3 \cos59 \degree × \cosec 31 \degree[/tex]
Solution :
[tex] \qquad \sf = 5 \sin50 \degree× \sec40 \degree-3 \cos59 \degree × \cosec 31 \degree[/tex]
[tex] \qquad \sf = 5 \sin50 \degree× \sec(90 - 50) \degree-3 \cos59 \degree × \cosec (90 - 59) \degree[/tex]
[tex] \qquad \sf = 5 \sin50 \degree× \cosec 50 \degree-3 \cos59 \degree × \sec59\degree[/tex]
[tex] \displaystyle\qquad \sf = 5 \: \cancel{ \sin50 }\degree× \frac{1}{ \cancel{\sin50}\degree} -3 \: \cancel{\cos59 }\degree × \frac{1}{ \cancel{\cos59}\degree} [/tex]
[tex] \displaystyle\qquad \sf = \: 5 - 3[/tex]
[tex] \red{ \displaystyle\qquad \bf = \: 2}[/tex]
Hence,
[tex] \red{ \sf \therefore \sf5 \sin50 \degree× \sec40 \degree-3 \cos59 \degree × \cosec 31 \degree = 2}[/tex]
[tex] \: [/tex]