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  • 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.(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=

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1)fracsec a + an a 1-sin a = frac1+sin acos a *(1- sin a)=frac1+sin^2 a+ 2 sin a cos a *(1-sin^2 a)

=sec^3 a +sec a * an^2 a+2 sec^2 a * an a

2) If ; an(a+b)=1; then ; sin (a+b)

Rightarrow an(a+b)=1

Rightarrow (a+b)= fracprod4

Rightarrow sin (a+b)=sin ( fracprod4)=frac1sqrt2

3) an 0 an 1 an2 ...... an 90

wkt ; ; tan 90 =infty

so, ; ; an 0 an 1 an2 ...... an 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 ; ;( 2prod*r) ; where ; ; r; is; ; radius ; ; of ; ; circle

4*a=2*prod *r

Rightarrow a=fracprod*r2

Area ; of; square = a^2=(fracprod*r2)^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

Posted by

Deependra Verma

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