prove that tan2 /1+tan2+cot2/1+cot2=sec×cosec-2sin×cos
Answers
Answer:
We have,
LHS = tan
2
θ+cot
2
θ+2
⇒ LHS = tan
2
θ+cot
2
θ+2tanθcotθ
⇒ LHS = (tanθ+cotθ)
2
⇒ LHS = (
cosθ
sinθ
+
sinθ
cosθ
)
2
⇒ LHS = (
sinθcosθ
sin
2
θ+cos
2
θ
)
2
⇒ LHS = (
sinθcosθ
1
)
2
⇒ LHS =
sin
2
θcos
2
θ
1
=cosec
2
θsec
2
θ=RHS
ALTERNATIVELY 1
We have,
LHS = tan
2
θ+cot
2
θ+2=(1+tan
2
θ)+(1+cot
2
θ)
⇒ LHS = sec
2
θ+cosec
2
θ=
cos
2
θ
1
+
sin
2
θ
1
=
cos
2
θsin
2
θ
sin
2
θ+cos
2
θ
⇒ LHS =
cos
2
θsin
2
θ
1
=cosec
2
θsec
2
θ=RHS
ALTERNATIVELY 2
LHS = tan
2
θ+cot
2
θ+2
⇒ LHS = 1 + tan
2
θ+cot
2
θ + 1
⇒ LHS = 1 + tan
2
θ+cot
2
θ+tan
2
+θcot
2
θ [∵tan
2
θcot
2
θ=1]
⇒ LHS = (1 + tan
2
θ) + cot
2
θ(1 + tan
2
θ)
⇒ LHS = (1 + tan
2
θ)(1 + cot
2
θ) = sec
2
θcosec
2
θ=RHS
ALTERNATIVELY 3
RHS = sec ^2
θ+cosec ^2θ
⇒ RHS = (1 + tan ^2θ)(1 + cot ^2θ)
⇒ RHS = 1 + tan ^2
θ+cot ^2
θ+tan ^2
θcot ^2 θ
⇒ RHS = 1 + tan ^2
θ+cot ^2
θ+1=tan ^2
θ+cot ^2
θ+2=LHS
Answer:
We have,
LHS = tan2θ+cot2θ+2
⇒ LHS = tan2θ+cot2θ+2tanθcotθ
⇒ LHS = (tanθ+cotθ)2
⇒ LHS = (cosθsinθ+sinθcosθ)2
⇒ LHS = (sinθcosθsin2θ+cos2θ)2
⇒ LHS = (sinθcosθ1)2
⇒ LHS = sin2θcos2θ1=cosec2θsec2θ=RHS