3CotA=6SecB=-2√10.Prove that:-CosecA-tanB=2
Answers
Answer:
Given 3 cotA = 4.
cotA = 4/3. ----------------- (1).
We know that tan theta = 1/cot theta
tan A = 1/cotA
= 1/4/3
= 3/4. --------------- (2)
We know that cosec^2 theta = 1 + cot^2 theta
= 1 + (4/3)^2
= 1 + 16/9
= 25/9.
cosec theta = 5/3. ---------------- (3)
We know that cosec theta = 1/sin theta
(5/3) = 1/sin theta
5sin theta = 3
sin theta = 3/5 ------------- (4)
We know that cos^2 theta = 1 - sin^2 theta
= 1 - (3/5)^2
= 1 - 9/25
= 25 - 9/25
= 16/25
cos A = 4/5. ------------- (5)
Now, LHS:
\frac{1 - tan^2A}{1 + tan^2A}
1+tan
2
A
1−tan
2
A
\frac{1 - (\frac{3}{4})^2 }{1 + ( \frac{3}{4})^2 }
1+(
4
3
)
2
1−(
4
3
)
2
\frac{1 - \frac{9}{16} }{1 + \frac{9}{16} }
1+
16
9
1−
16
9
\frac{16 - 9}{16 + 9}
16+9
16−9
\frac{7}{25}
25
7
.
Now, RHS: (4)&(5)
cos^2A - sin^2Acos
2
A−sin
2
A
( \frac{4}{5})^2 - ( \frac{3}{5} )^2(
5
4
)
2
−(
5
3
)
2
\frac{16}{25} - \frac{9}{25}
25
16
−
25
9
\frac{16-9}{25}
25
16−9
\frac{7}{25}
25
7
LHS = RHS.