Prove the identity.cscx+1cot2x=cscx−1Note that each Statement must be based on a Rule chosen from the Rule menu. To se the right of the Rule.Statement
Q. Prove the identity.cscx+1cot2x=cscx−1Note that each Statement must be based on a Rule chosen from the Rule menu. To se the right of the Rule.Statement
Start with left-hand side: We start with the left-hand side of the identity and try to transform it into the right-hand side.We have: cscx+1cot2xRecall the trigonometric identities: cot(x)=tan(x)1 and csc(x)=sin(x)1
Rewrite cot(x): Rewrite cot(x) in terms of sin(x) and cos(x): cot(x)=sin(x)cos(x)So, cot2(x)=(sin(x)cos(x))2
Rewrite left-hand side: Now, let's rewrite the left-hand side using the cot(x) in terms of sin(x) and cos(x):cscx+1cot2x=sin(x)1+1(sin(x)cos(x))2
Multiply by sin(x): Multiply both numerator and denominator by sin(x) to get rid of the fraction in the denominator:(sin(x)cos(x))2⋅(sin(x)1+1)⋅sin(x)sin(x)=sin2(x)cos2(x)⋅1+sin(x)sin(x)
Simplify expression: Simplify the expression by canceling out a sin(x) in the numerator and denominator: sin(x)cos2(x)÷(1+sin(x)) = sin(x)⋅(1+sin(x))cos2(x)
Use Pythagorean identity: Use the Pythagorean identity: sin2(x)+cos2(x)=1 So, cos2(x)=1−sin2(x) Replace cos2(x) in the expression: (1−sin2(x))/(sin(x)⋅(1+sin(x)))
Factor numerator: Factor the numerator as a difference of squares: sin(x)⋅(1+sin(x))(1−sin(x))⋅(1+sin(x))
Cancel common factor: Cancel out the common factor (1+sin(x)) in the numerator and denominator:sin(x)1−sin(x)
Rewrite sin(x)1: Rewrite sin(x)1 as csc(x):csc(x)−sin(x)sin(x)= csc(x)−1
Transform into right-hand side: We have now transformed the left-hand side of the identity into the right-hand side:cscx+1cot2x=cscx−1This completes the proof.