cosec (65+ A) - sec (25" - A)
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Answer:
⇒ cosec(65° + θ) – sec(25° – θ) + tan220° - cosec2 70°
⇒ cosec(65 + θ) – cosec[90 – (25 – θ) + tan220 – sec2(90 – 70)
⇒ cosec (65 + θ) – cosec(65 + θ) + tan220 – sec220 = (-1)
∴ tan2 θ – sec2 θ = (-1)
Alternate Method For angles A and B
(A + B) = 90°
⇒ cosec A = sec B ----(1)
According to question
⇒ cosec(65° + θ) – sec(25° – θ) + tan220° - cosec2 70°
⇒ cosec(65° + θ) – cosec(90° - (25° – θ) + sec220° - 1 - cosec2 70°
Using eqn (1)
⇒ cosec(65° + θ) – cosec(65° – θ) + sec220° - 1 - cosec2 70°
⇒ -1
Verified answer
Answer:
[tex]\boxed{\bf\: cosec( {65}^{ \circ} + A) - sec(25^{ \circ} - A) = 0 \: } \\ [/tex]
Step-by-step explanation:
Given expression is
[tex]\sf\: cosec( {65}^{ \circ} + A) - sec(25^{ \circ} - A) \\ [/tex]
can be rewritten as
[tex]\sf\: = \: cosec( {65}^{ \circ} + A) - sec(90^{ \circ} - 65^{ \circ} - A) \\ [/tex]
[tex]\sf\: = \: cosec( {65}^{ \circ} + A) - sec\left[ 90^{ \circ} -( 65^{ \circ} + A)\right] \\ [/tex]
We know,
[tex]\boxed{\sf\:sec(90^{ \circ} - x) = cosecx \: } \\ [/tex]
So, Using this result, we get
[tex]\sf\: = \: cosec( {65}^{ \circ} + A) -cosec( 65^{ \circ} + A)\\ [/tex]
[tex]\sf\: = \: 0\\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: cosec( {65}^{ \circ} + A) - sec(25^{ \circ} - A) = 0 \: } \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[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]