#cos⁻¹(sqrtcos α)−tan⁻¹(sqrtcos α)=x# ,then what is the value of sin x ?

Question

(a) #tan^2(α/2)#
(b) #cot^2(α/2)#
(c) #tanα#
(d) #cot (α/2)#

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Answer 1 · 2018-05-15T16:10:28

#sinx=tan(alpha/2)-cosalpha/(sqrt2cos(alpha/2))#

Let #sqrtcosalpha=m#

#rarrcos^(-1)(m)-tan^(-1)(m)=x#

Let #cos^(-1)m=y# then #cosy=m#

#rarrsiny=sqrt(1-cos^2y)=sqrt(1-m^2)#

#rarry=sin^(-1)(sqrt(1-m^2))=cos^(-1)m#

Also, let #tan^(-1)m=z# then #tanz=m#

#rarrsinz=1/cscz=1/sqrt(1+cot^2z)=1/sqrt(1+(1/m)^2)=m/sqrt(1+m^2)#

#rarrz=sin^(-1)(m/sqrt(1+m^2))=tan^(-1)m#

#rarrcos^(-1)(m)-tan^(-1)(m)#

#=sin^(-1)(sqrt(1-m^2))-sin^(-1)(m/sqrt(1+m^2))#

#=sin^-1(sqrt(1-m^2)*sqrt(1-(m/sqrt(1+m^2))^2)-(m/sqrt(1+m^2))*sqrt(1-(sqrt(1-m^2))^2))#

#=sin^(-1)(sqrt((1-cosalpha)/(1+cosalpha))-cosalpha/sqrt(1+cosalpha))#

#=sin^(-1)(tan(alpha/2)-cosalpha/(sqrt2cos(alpha/2)))=x#

#rarrsinx=sin(sin^(-1)(tan(alpha/2)-cosalpha/(sqrt2cos(alpha/2))))=tan(alpha/2)-cosalpha/(sqrt2cos(alpha/2))#