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Question Number 62347 by rajesh4661kumar@gamil.com last updated on 20/Jun/19

Answered by Kunal12588 last updated on 20/Jun/19

((csc θ − sin θ)/(sec θ − cos θ)) = (a^3 /b^3 )  ⇒cot θ ((1−sin^2 θ)/(1−cos^2 θ))=(a^3 /b^3 )  ⇒cot θ=(a/b)  ⇒cot^2 θ=(a^2 /b^2 )  ⇒1+cot^2 θ=((a^2 +b^2 )/b^2 )  ⇒csc^2 θ=((a^2 +b^2 )/b^2 )   ((csc^2 θ−1)/(csc θ))=a^3   ⇒(((a^2 +b^2 −b^2 )/b^2 )/((√(a^2 +b^2 ))/b))=a^3   ⇒a^2 =a^3 b(√(a^2 +b^2 ))  ⇒1=ab(√(a^2 +b^2 ))  ⇒a^2 b^2 (a^2 +b^2 )=1

cscθsinθsecθcosθ=a3b3cotθ1sin2θ1cos2θ=a3b3cotθ=abcot2θ=a2b21+cot2θ=a2+b2b2csc2θ=a2+b2b2csc2θ1cscθ=a3a2+b2b2b2a2+b2b=a3a2=a3ba2+b21=aba2+b2a2b2(a2+b2)=1

Commented by peter frank last updated on 20/Jun/19

nice

nice

Answered by Kunal12588 last updated on 20/Jun/19

ANOTHER WAY  cosec θ − sin θ = a^3   ⇒_(tan^(−1)   t = θ)  ((√(1+t^2 ))/t)−(t/(√(1+t^2 )))=a^3   ⇒1+t^2 −t^2 =a^3 t(√(1+t^2 ))  ⇒(1/(a^3 t))=(√(1+t^2 ))  sec θ − cos θ =b^3   ⇒_(tan^(−1)  t = θ)  (√(1+t^2 ))−(1/(√(1+t^2 )))=b^3   ⇒1+t^2 −1=b^3 (√(1+t^2 ))  ⇒(t^2 /b^3 )=(√(1+t^2 ))  (1/(a^3 t))=(t^2 /b^3 )  ⇒t=(b/a)  putting the value of  t  ((b^2 /a^2 )/b^3 )=(√(1+(b^2 /a^2 )))  ⇒(1/(a^2 b))=((√(a^2 +b^2 ))/a)  ⇒ab(√(a^2 +b^2 ))=1  ⇒a^2 b^2 (a^2 +b^2 )=1

ANOTHERWAYcosecθsinθ=a3tan1t=θ1+t2tt1+t2=a31+t2t2=a3t1+t21a3t=1+t2secθcosθ=b3tan1t=θ1+t211+t2=b31+t21=b31+t2t2b3=1+t21a3t=t2b3t=baputtingthevalueoftb2a2b3=1+b2a21a2b=a2+b2aaba2+b2=1a2b2(a2+b2)=1

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