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The general solution is ##npi+-pi/3## As ##sec^2x-2=0## we have ##sec^2x-(sqrt2)^2=0## or ##(secx+sqrt2)(secx-sqrt2)=0## and ##secx=+-sqrt2=+-sec(pi/3)## Hence in the range ##(-pi+pi)## the values are ##{+-pi/3+-(2pi)/3}## which can also […]

Please refer to explanation below Start from the left hand side ##(sinx+cosx)^4 = (1+2sinx cosx)^2## ##(sinx+cosx)(sinx+cosx)]^2## Expand/ multiply/foil the expression ##(sin^2x + sinxcosx+sinxcosx+cos^2x)^2## Combine like terms […]

##cos(2pi/5)=(-1+sqrt(5))/4## Here the most elegant solution I found in: http://math.stackexchange.com/questions/7695/how-to-prove-cos-frac2-pi-5-frac-1-sqrt54 ##cos(4pi/5)=cos(2pi-4pi/5)=cos(6pi/5)## So if ## x=2pi/5##: ##cos(2x)=cos(3x)## Replacing the cos(2x) and cos(3x) by their general formulae: ##color(red)(cos(2x)=2cos^2x-1 […]

Both pairs of opposite sides are parallel

-3/2 240 degrees which is in the third quadrant has a cos value of -cos60 degrees = -1/2 and cos 0 degrees = 1. Thus -1/2 […]

##(3pi)/4## radians ##tan(pi-pi/4)=tan(3pi/4)## = -1. Hence ##arctan(-1)=(3pi)/4## radians

##(8/38/3)## The ##x##-coordinate of the centroid is simply the average of the ##x##-coordinates of the triangle’s vertices. The same logic is applied to the ##y##-coordinates for […]

Option( c)is correct i.e.AB=BC+AD Given ##In quadrilateral ABCD /_DAB=/_ABC=60^@## ##/_CAB=/_CBD## Construction ## BC and AD are produced to intersect at E## ## In Delta ABE/_EBA=/_EAB=60^@ =>/_BEA=60^@## […]

AAS – angle angle side SSA – side side angle SAS – side angle side SSS- side side side