prove that cos sq a/2+cos sq b/2+cos sq c/2=2+2sin a/2+sin b/2+ sin c/2
Answers
LHS
=(sinA+cosecA)
2
+(cosA+secA)
2
=sin
2
A+cosec
2
A+2sinA.cosec A+cos
2
A+sec
2
A+2cosA.secA
=sin
2
A+cos
2
A+2+2+cosec
2
A+sec
2
A
=5+(1+cot
2
A)+(1+tan
2
A) ........ [sin²θ+cos²θ=1], [sec
2
θ=1+tan
2
θ] and [cosec
2
x=1+cot
2
x]
=7+tan
2
A+cot
2
A
= RHS
Hence, proved that
(sinA+cosec A)
2
+(cosA+secA)
2
=7+tan
2
A+cot
2
A
~~~NEW FOLLOWERS WANTED!!~~~
Answer:
this is your Answer hope it helps you
Step-by-step explanation:
Given u=
a
2
cos
2
θ+b
2
sin
2
θ
+
a
2
sin
2
θ+b
2
cos
2
θ
∴u
2
=(a
2
+b
2
)cos
2
θ+(a
2
+b
2
)sin
2
θ+2
a
2
cos
2
θ+b
2
sin
2
θ
a
2
sin
2
θ+b
2
cos
2
θ
u
2
=a
2
+b
2
+2
a
4
cos
2
θsin
2
θ+b
4
cos
2
θsin
2
θ+a
2
b
2
(cos
4
θ+sin
4
θ)
=a
2
+b
2
+2
(a
4
+b
4
)cos
2
θsin
2
θ+a
2
b
2
(1−2cos
2
θsin
2
θ)
u
2
=a
2
+b
2
+2
4
a
4
+b
4
(sin2θ)
2
+a
2
b
2
(1−
2
sin
2
2θ
)
u
2
=a
2
+b
2
+2
4
(a
2
−b
2
)
2
(sin2θ)
2
+a
2
b
2
For max. value of u
2
, sin
2
2θ=1
u
2
=a
2
+b
2
+2
2
(a
2
+b
2
)
2
=2(a
2
+b
2
)
And for min. u
2
,sin2θ=0
u
2
=a
2
+b
2
+2ab=(a+b)
2
∴ Difference between max. and min. value of u
2
is 2(a
2
+b
2
)−(a+b)
2
=(a−b)
2
.