In acute angled triangle, $\cos \frac{A}{2} > \sin \frac{A}{2}$
So, $\sin \frac{A}{2}.\cos \frac{A}{2} > \sin \frac{A}{2}.\sin \frac{A}{2}$
Hence, $2\sin \frac{A}{2}\cos \frac{A}{2} > 2{\sin ^2}\frac{A}{2}$
Thus, $\sin A > 1 – \cos A$
Likewise, $\sin B > 1 – \cos B$ & $\sin C > 1 – \cos C$ Continue reading In acute angled $\Delta ABC$ prove that,
$sin A + sin B + sin C > cos A + cos B + cos C$
The area of triangle OAB bounded by the tangent to $xy=c^2$ in the 1st quadrant, x-axis & y-axis is (refer figure):
(1) Maximum if P is the midpoint of AB
(2) Increases as P moves downwards or upwards
(3) Constant
(4) Independent of c
Solution
$xy=c^2$ is rectangular hyperbola. The equation of tangent in parametric form at some point P $(ct, \frac {c}{t})$ is given by,
$\frac {x}{t}+yt=2c$
At point A, $x=2ct=OA$
At point B, $y=\frac {2c}{t}=OB$
Area of $\Delta OAB$ = $\frac {1}{2}.OA.OB$ = $\frac {1}{2}.2ct.\frac {2c}{t}$ = $2c^2$
Hence, Option (3).
The equation of the plane containing the line 2x-5y+z=3, x+y+4z=5 and parallel to the plane x+3y+6z=1 is
(1) x+3y+6z=-7
(2) x+3y+6z=7
(3) 2x+6y+12z=-13
(4) 2x+6y+12z=13
We have, $8(1 – {\cos ^2}x) + 2(4{\cos ^3}x – 3\cos x) + 6\cos x = 7$
$ \Rightarrow 8{\cos ^3}x – 8{\cos ^2}x + 1 = 0$
Let $cos x = t$
So, $8{t^3} – 8{t^2} + 1 = 0$ Continue reading Solve for $x \in \left[ {0,\frac{\pi }{2}} \right]$
$8{\sin ^2}x + 2\cos 3x + 6\cos x = 7$
A ladder of length $l=2\sqrt 6$ unit is resting against a wall just touching the cube of edge 1 unit as shown in the figure. Find h above the cube.
Solution
We have, $\tan \theta = h$ and $\sin \theta = \frac{{h + 1}}{l}$
So, $2\sqrt 6 \sin \theta = 1 + \tan \theta $ Continue reading Ladder, Cube & Wall Problem
$\begin{array}{l}y = \frac{{{x^4} – {x^2}}}{{{x^6} + 2{x^3} – 1}},x > 1\\{y_{\max }} = ?\end{array}$
The given integral =$ – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{x(1 + x)}}dx} $
$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{1 + x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $
$ = \int\limits_{ – 1}^0 {\frac{{\ln \left[ {1 + ( – 1 – x) + {{( – 1 – x)}^2}} \right]}}{{1 + (-1-x)}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $
$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{ – x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $
$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $
$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln \frac{{(1 – x)(1 + x + {x^2})}}{{(1 – x)}}}}{x}dx} $ Continue reading Prove that,
$\int\limits_0^1 {\frac{1}{{x(1 + x)}}.\ln \left[ {\frac{1}{{1 + x(1 + x)}}} \right]dx} = $
$\frac{4}{3}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ……..\infty \, terms} \right)$
