Prove that
coseco
cota (90°-o) by
tan²o-1
+cosec²o by
sin2o-cosec2o=1 by sin2o-cos2o
Answers
A=0 finally thanks
Step-by-step explanation:
Answer:
\frac{tan(90-A)cotA}{cosec^{2}A} - cos^{2}A=0
cosec
2
A
tan(90−A)cotA
−cos
2
A=0
Step-by-step explanation:
LHS = \frac{tan(90-A)cotA}{cosec^{2}A} - cos^{2}ALHS=
cosec
2
A
tan(90−A)cotA
−cos
2
A
=\frac{cotA\cdot cotA}{cosec^{2}A} - cos^{2}A=
cosec
2
A
cotA⋅cotA
−cos
2
A
/* tan(90-A) = cotA */
= \frac{cot^{2}A}{cosec^{2}A}-cos^{2}A=
cosec
2
A
cot
2
A
−cos
2
A
=\frac{\frac{cos^{2}A}{sin^{2}A}}{\frac{1}{sin^{2}A}}-cos^{2}A=
sin
2
A
1
sin
2
A
cos
2
A
−cos
2
A
________________________
\begin{gathered}i)cotA=\frac{cosA}{sinA}\\ii) cosecA = \frac{1}{sinA}\end{gathered}
i)cotA=
sinA
cosA
ii)cosecA=
sinA
1
________________________
\begin{gathered}=\cos^{2}A-cos^{2}A\\=0\\=RHS\end{gathered}
=cos
2
A−cos
2
A
=0
=RHS
Therefore,
\frac{tan(90-A)cotA}{cosec^{2}A} - cos^{2}A=0
cosec
2
A
tan(90−A)cotA
−cos
2
A=0