Please solve the attached question
Answers
Answer:
IDENTITY USED :-
★ cos2A = cos²A - sin²A
★ 1 = sin²A + cos²A
★ cosA/sinA = cotA
\begin{gathered} \\ \end{gathered}
SOLUTION :-
\begin{gathered} \\ \tt \: L.H.S = \dfrac{1 + cos2A}{1 - cos2A} \\ \\ \\ \bigstar\boxed{ \sf \: cos2A = {cos}^{2}A - {sin}^{2}A } \\ \\ \\ \tt \implies \: \dfrac{1 + ( {cos}^{2}A - {sin}^{2}A) }{1 - {(cos}^{2}A - {sin}^{2}A )} \\ \\ \\ \bigstar \boxed{ \sf \: 1 = {sin}^{2}A + {cos}^{2}A } \\ \\ \\ \tt \implies \: \dfrac{ \cancel{{sin}^{2}A} + {cos}^{2}A + {cos}^{2}A - \cancel {{sin}^{2}A }}{ {sin}^{2}A + \cancel{{cos}^{2}A }- \cancel {{cos}^{2}A} + {sin}^{2}A } \\ \\ \\ \tt \implies \: \dfrac{2 {cos}^{2}A }{2 {sin}^{2}A } \\ \\ \\ \bigstar \boxed{ \sf \: \dfrac{cosA}{sinA} = cotA} \\ \end{gathered}
L.H.S=
1−cos2A
1+cos2A
★
cos2A=cos
2
A−sin
2
A
⟹
1−(cos
2
A−sin
2
A)
1+(cos
2
A−sin
2
A)
★
1=sin
2
A+cos
2
A
⟹
sin
2
A+
cos
2
A
−
cos
2
A
+sin
2
A
sin
2
A
+cos
2
A+cos
2
A−
sin
2
A
⟹
2sin
2
A
2cos
2
A
★
sinA
cosA
=cotA
\begin{gathered} \\ \implies \tt \: {cot}^{2}A = R.H.S \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (verified)\end{gathered}
⟹cot
2
A=R.H.S(verified)
\begin{gathered} \\ \end{gathered}