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##Cos2x## = ##Cos2x## ##Cos2x## = ##(1 – tan^2x)/(1+tan^2x)## ##Cos2x## = ##(1 – tan^2x)/(Sec^2x)## ##Cos2x## = ##(1 – tan^2x)/(1/Cos^2x)## ##Cos2x## = ##(1 – sin^2x/Cos^2x)*(Cos^2x)## ##Cos2x## = ##Cos^2x […]

Proof We know that ##color(blue)(cottheta = cos theta/sin theta## Left Hand Side = ##sintheta*cot theta## = ##cancel(sintheta)*(cos theta/cancel(sintheta))## = ##cos theta## = Right Hand Side Hence […]

See the derivation in the Explanation Section below. Suppose that ##(30+theta)=alpha & (60-theta)=beta## so that ##alpha+beta=90 or beta=90-alpha##. Sub.ing these in the given eqn. we have […]

##{ pi/2 + 2kpi k in Z }## Recall ##tanx = sinx/cosx## then ##tan4x = (sin4x)/(cos4x) and tan2x = (sin2x)/(cos2x)## ##(sin4x)/(cos4x) = (sin2x)/(cos2x)## [Cross multiplying] ##sin4xcos2x […]

I got ##59.45^@## (using Cosine formula …see A.S.Adikesavan’s answer using Sine formula) My understanding is that: ##color(blue)——————————————————-## ##color(blue)(| )Given vectors: ## ##color(blue)(| )color(white)(XXX)color(blue)( vecu=< u_1u_2u_3 > […]

##csc^-1 (-1)=-pi/2##. Recall the Defn. of ##csc^-1## function : ##csc^-1 x=theta |x| ge 1 iff csc theta=x theta in [-pi/2pi/2]-{0}##. We have ##csc(-pi/2)=-1 where -pi/2 in […]

##pi/2< (7pi)/9 < pi## So the reference angle for ##(7pi)/9## is ##pi - (7pi)/9## ##=(2pi)/9##

40 degrees Since the two angles are compliments they must add up to 90. Call the angles ##x## and ##y## Then ##x+ y = 90## Since […]

##sinx/(1-cosx)+(1-cosx)/sinx = 2cscx## ##sinx/(1-cosx)+(1-cosx)/sinx = [sin^2x+(1-cosx)^2]/[(1-cosx)sinx]## ## = [sin^2x+cos^2x+1-2cosx]/[(1-cosx)sinx]## ## = [1+1-2cosx]/[(1-cosx)sinx]## ## = [2-2cosx]/[(1-cosx)sinx]## ## = 2cancel[1-cosx]/[cancel(1-cosx)sinx]## ## = 2/sinx## ## = 2cscx##