prove that sinA + cos A/sin A - cos A + sinA - cos A /sin A - cos A
Answers
Step-by-step explanation:
+
sin(a)+cos(a)
sin(a)−cos(a)
=
1−cos(a)
2
2
Solution :-
\sf \implies \dfrac{ \sin(a) + \cos(a) }{ \sin(a) - \cos(a) } + \dfrac{ \sin(a) - \cos(a) }{ \sin(a) + \cos(a) } = \dfrac{2}{1 - { \cos(a) }^{2} }⟹
sin(a)−cos(a)
sin(a)+cos(a)
+
sin(a)+cos(a)
sin(a)−cos(a)
=
1−cos(a)
2
2
Taking LHS :-
\sf \implies \dfrac{ \sin(a) + \cos(a) }{ \sin(a) - \cos(a) } + \dfrac{ \sin(a) - \cos(a) }{ \sin(a) + \cos(a) }⟹
sin(a)−cos(a)
sin(a)+cos(a)
+
sin(a)+cos(a)
sin(a)−cos(a)
\sf \implies \dfrac{( \sin(a) + \cos(a) {)}^{2} + ( { \sin(a) - \cos(a) })^{2} }{ \sin(a)^{2} - \cos(a)^{2} }⟹
sin(a)
2
−cos(a)
2
(sin(a)+cos(a))
2
+(sin(a)−cos(a))
2
\sf \implies \: \dfrac{ { {\sin(a)}^{2} + { \cos(a) }^{2} + 2 \sin(a) \cos(a) + { \sin(a) }^{2} + { \cos(a) }^{2} - 2 \sin(a) \cos(a) }^{2} }{ { \sin(a) }^{2} - { \cos(a) }^{2} }⟹
sin(a)
2
−cos(a)
2
sin(a)
2
+cos(a)
2
+2sin(a)cos(a)+sin(a)
2
+cos(a)
2
−2sin(a)cos(a)
2
\sf \implies \: we \: know \: that \: { \sin(a) }^{2} + { \cos(a) }^{2} = 1⟹weknowthatsin(a)
2
+cos(a)
2
=1
\sf \implies \: \dfrac{ { 1 + 2 \sin(a) \cos(a) + 1 - 2 \sin(a) \cos(a) }}{ { \sin(a) }^{2} - { \cos(a) }^{2} }⟹
sin(a)
2
−cos(a)
2
1+2sin(a)cos(a)+1−2sin(a)cos(a)
\sf \implies \: \dfrac{ { 2 + 2 \sin(a) \cos(a) - 2 \sin(a) \cos(a) } }{ { \sin(a) }^{2} - { \cos(a) }^{2} }⟹
sin(a)
2
−cos(a)
2
2+2sin(a)cos(a)−2sin(a)cos(a)
\sf \implies \: \dfrac{ { 2 + \cancel{2 \sin(a) \cos(a) } \cancel {- 2 \sin(a) \cos(a) } } }{ { \sin(a) }^{2} - { \cos(a) }^{2} }⟹
sin(a)
2
−cos(a)
2
2+
2sin(a)cos(a)
−2sin(a)cos(a)
\sf \implies \: \dfrac{ { 2 }}{ { \sin(a) }^{2} - { \cos(a) }^{2} }⟹
sin(a)
2
−cos(a)
2
2
\sf \implies \: we \: know \: that \: { \sin(a) }^{2} =1 - { \cos(a) }^{2}⟹weknowthatsin(a)
2
=1−cos(a)
2
\sf \implies \: \dfrac{ { 2 }}{ 1 - { \cos(a) }^{2} - { \cos(a) }^{2} }⟹
1−cos(a)
2
−cos(a)
2
2
\sf \implies \: \dfrac{ { 2 }}{ 1 - 2 { \cos(a) }^{2} }⟹
1−2cos(a)
2
2