Question

answer the following questions

Answer 1

Consider L.H.S:
$\ \ \sqrt(\frac{1-\sin A}{1+\sin A}) \\=\sqrt(\frac{1-\sin A}{1+\sin A}) \sqrt(\frac{1-\sin A}{1-\sin A})\\= \frac{1-\sin A}{\sqrt(1-\sin ^2 A)} \\= \frac{1-\sin A}{\cos A} \\= \sec A - \tan A$
This is equal to R.H.S.

Answer 2

$A. \sqrt{ \frac{1-sinA}{1+sinA} } \\ \\ =\sqrt{ \frac{1-sinA}{1+sinA} \times \frac{1-sinA}{1-sinA} } \\ \\ =\sqrt{ \frac{(1-sinA) \times (1-sinA)}{(1+sinA) \times (1+sinA)} } \\ \\ = \sqrt{ \frac{(1-sinA)^2}{cos^2A} }$

$= \frac{(1-sinA)}{cosA} } \\ \\ = \frac{1}{cosA} - \frac{sinA}{cosA} \\ \\ =secA-tanA\ \ \ \ \ (proved)$

$From\ \frac{1}{cosA} - \frac{sinA}{cosA} = secA-secA.sinA=secA(1-sinA)\ \ \ \ (proved)$

B. 
marks           no.of students          cumulative
0-10                     7                            7
10-20                   6                       7+6= 13
20-30                   5                       13+5=18
30-40                   8                       18+8=26
40-50                   9                        26+9=35

(35+1)/2 = 18
So median group = 20-30
L = 20
n = 35
cf = 13
f = 5
w = 10

median = $L+ \frac{n/2-cf}{f} \times w=20+ \frac{17.5-13}{5} \times 10=29$

median is 29