Eliminate 'x' in:
3secx - tanx = a
secx + 2tanx = b
Answers
Answer:
∫
1+2tanx(tanx+secx)
=−logcosx+log(secx+tanx)+C
Step-by-step explanation:
Given:
We have,
\int\sqrt{1+2tanx(tanx+secx)}∫
1+2tanx(tanx+secx)
= \int\sqrt{1+2tan^2x+2tanxsecx}dx∫
1+2tan
2
x+2tanxsecx
dx
= \int \sqrt{1 +2\frac{sin^2x}{cos^2x}+2 \times\frac{sinx}{cosx} \times \frac{1}{cosx}} dx∫
1+2
cos
2
x
sin
2
x
+2×
cosx
sinx
×
cosx
1
dx [tanx=\frac{sinx}{cosx}, secx=\frac{1}{cosx} ][tanx=
cosx
sinx
,secx=
cosx
1
]
= \int \sqrt{1 +2\frac{sin^2x}{cos^2x}+2 \frac{sinx}{cos^2x}} dx∫
1+2
cos
2
x
sin
2
x
+2
cos
2
x
sinx
dx
= \int \sqrt{\frac{cos^2x+sin^2x+sin^2x+2sinx}{cos^2x} }dx∫
cos
2
x
cos
2
x+sin
2
x+sin
2
x+2sinx
dx
Using the trigonometric identity cos^2x+sin^2x=1cos
2
x+sin
2
x=1
= \int \sqrt{\frac{1+sin^2x+2sinx}{cos^2x} }dx∫
cos
2
x
1+sin
2
x+2sinx
dx
= \int \sqrt{\frac{sinx+1}{cos^2x} }dx∫
cos
2
x
sinx+1
dx [(sinx+1)^2=1+sin^2x+2sinx][(sinx+1)
2
=1+sin
2
x+2sinx]
= \int(\frac{sinx+1}{cosx})dx∫(
cosx
sinx+1
)dx
= \int \frac{sinx}{cosx}+ \frac{1}{cosx} dx∫
cosx
sinx
+
cosx
1
dx
= \int(tanx+secx)dx∫(tanx+secx)dx
= - log cosx +log (secx+tanx) + C
Therefore,
\int\sqrt{1+2tanx(tanx+secx)} = - log cosx +log (secx+tanx) + C∫
1+2tanx(tanx+secx)
=−logcosx+log(secx+tanx)+C
Answer:
3a^2-8b^2+10ab=49.
Step-by-step explanation:
3secx-tanx=a
sex+2tanx=b
6secx-2tanx=2a
sex+2tanx=b
add both equation
7secx=2a+b
3(2a+b)/7 - tanx=a
tanx =6a+3b/7 - a
tanx= 3b-a/7
we know that
sec^2x -tan^2x=1
(2a+b/7)^2-(3b-a/7)^2=1
on solving you will get
3a^2-8b^2+10ab=49.