# 1.(sec A +Tan A)(1-Sin A)= 2.If Tan( A+B)=1 then Sin(A+B)= 3.Tan 0.Tan 1.Tan 2.........Tan 90= 4.perimeter of square and circumference of circle are equal then...........is more 5.sin 0 .sin 10. sin 30. sin 80. sin 90=

$1)\frac{\sec a +\tan a }{1-\sin a }= \frac{1+\sin a}{\cos a *(1- \sin a)}=\frac{1+\sin^{2} a+ 2 \sin a }{\cos a *(1-\sin^{2} a)}$

$=\sec^{3} a +\sec a * \tan^{2} a+2 \sec^{2} a * \tan a$

$2) If \; \tan(a+b)=1\; then \; sin (a+b)$

$\Rightarrow \tan(a+b)=1$

$\Rightarrow (a+b)= \frac{\prod}{4}$

$\Rightarrow \sin (a+b)=\sin ( \frac{\prod}{4})=\frac{1}{\sqrt{2}}$

$3) \tan 0 \tan 1 \tan2 ...... \tan 90$

$wkt \; \; tan 90 =\infty$

$so, \; \; \tan 0 \tan 1 \tan2 ...... \tan 90=\infty$

$4.perimeter of square and circumference of circle are equal then...........is more$

$perimeter\; \; of\; \; square \; \; is\; \; 4a \; \; where\; \; a \; \; is\; \; side\; \; of \; \; square$

$circumference\; \; of \; \; circle\; \; is \; \;( 2\prod*r) \; where \; \; r\; is\; \; radius \; \; of \; \; circle$

$4*a=2*\prod *r$

$\Rightarrow a=\frac{\prod*r}{2}$

$Area \; of\; square = a^2=(\frac{\prod*r}{2})^2=2.467*r^2$

$Area \; of\; circle = \prod*r^2=\prod*r^2=3.14*r^2$

So, Area of circle is greater than Area of square

$5)\sin 0* \sin 10*\sin 30*\sin 80*\sin 90=$

$we\; know\; that\; \sin 0 = 0$

$so \; \sin 0* \sin 10*\sin 30*\sin 80*\sin 90=0$

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