Here are the solutions for problems 1 to 12.
: Prove 1+cosxsinx+sinx1+cosx=sinx2
Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator.
LHS=(1+cosx)sinxsin2x+(1+cosx)2
Step 2: Expand the numerator.
LHS=(1+cosx)sinxsin2x+1+2cosx+cos2x
Step 3: Use the identity sin2x+cos2x=1.
LHS=(1+cosx)sinx1+1+2cosx=(1+cosx)sinx2+2cosx
Step 4: Factor out 2 from the numerator and cancel the common term.
LHS=(1+cosx)sinx2(1+cosx)=sinx2
This matches the right-hand side (RHS).
\frac{2{\sin x}}
: Prove 1−cosxsinx=sinx1+cosx
Step 1: Start with the left-hand side (LHS) and multiply the numerator and denominator by the conjugate of the denominator, (1+cosx).
LHS=1−cosxsinx×1+cosx1+cosx
Step 2: Multiply the terms in the numerator and denominator.
LHS=12−cos2xsinx(1+cosx)
Step 3: Use the identity 1−cos2x=sin2x.
LHS=sin2xsinx(1+cosx)
Step 4: Cancel out sinx from the numerator and denominator.
LHS=sinx1+cosx
This matches the right-hand side (RHS).
\frac{1+\cos x{\sin x}}
: Prove 1−cosxsinx−sinx1−cosx=2cotx
Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator.
LHS=(1−cosx)sinxsin2x−(1−cosx)2
Step 2: Expand the numerator.
LHS=(1−cosx)sinxsin2x−(1−2cosx+cos2x)=(1−cosx)sinxsin2x−1+2cosx−cos2x
Step 3: Use the identity sin2x−1=−cos2x.
LHS=(1−cosx)sinx−cos2x+2cosx−cos2x=(1−cosx)sinx2cosx−2cos2x
Step 4: Factor out 2cosx from the numerator.
LHS=(1−cosx)sinx2cosx(1−cosx)
Step 5: Cancel out the common term (1−cosx).
LHS=sinx2cosx=2cotx
This matches the right-hand side (RHS).
2cotx
: Prove cosx1+sinx+1+sinxcosx=cosx2
Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator.
LHS=cosx(1+sinx)(1+sinx)2+cos2x
Step 2: Expand the numerator.
LHS=cosx(1+sinx)1+2sinx+sin2x+cos2x
Step 3: Use the identity sin2x+cos2x=1.
LHS=cosx(1+sinx)1+2sinx+1=cosx(1+sinx)2+2sinx
Step 4: Factor out 2 from the numerator and cancel the common term.
LHS=cosx(1+sinx)2(1+sinx)=cosx2
This matches the right-hand side (RHS).
\frac{2{\cos x}}
: Prove cosx1+sinx−1+sinxcosx=2tanx
Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator.
LHS=cosx(1+sinx)(1+sinx)2−cos2x
Step 2: Expand the numerator.
LHS=cosx(1+sinx)1+2sinx+sin2x−cos2x
Step 3: Use the identity cos2x=1−sin2x.
LHS=cosx(1+sinx)1+2sinx+sin2x−(1−sin2x)=cosx(1+sinx)1+2sinx+sin2x−1+sin2x
Step 4: Simplify the numerator.
LHS=cosx(1+sinx)2sinx+2sin2x
Step 5: Factor out 2sinx from the numerator and cancel the common term.
LHS=cosx(1+sinx)2sinx(1+sinx)=cosx2sinx
Step 6: Express the result in terms of tanx.
LHS=2tanx
This matches the right-hand side (RHS).
2tanx
: Prove 1+sinxcosx+tanx=secx
Step 1: Start with the left-hand side (LHS) and express tanx as cosxsinx.
LHS=1+sinxcosx+cosxsinx
Step 2: Combine the fractions using a common denominator.
LHS=(1+sinx)cosxcos2x+sinx(1+sinx)
Step 3: Expand the numerator.
LHS=(1+sinx)cosxcos2x+sinx+sin2x
Step 4: Use the identity sin2x+cos2x=1.
LHS=(1+sinx)cosx1+sinx
Step 5: Cancel out the common term (1+sinx).
LHS=cosx1=secx
This matches the right-hand side (RHS).
secx
: Prove 1−sinxcosx−tanx=secx
Step 1: Start with the left-hand side (LHS) and express tanx as cosxsinx.
LHS=1−sinxcosx−cosxsinx
Step 2: Combine the fractions using a common denominator.
LHS=(1−sinx)cosxcos2x−sinx(1−sinx)
Step 3: Expand the numerator.
LHS=(1−sinx)cosxcos2x−sinx+sin2x
Step 4: Use the identity sin2x+cos2x=1.
LHS=(1−sinx)cosx1−sinx
Step 5: Cancel out the common term (1−sinx).
LHS=cosx1=secx
This matches the right-hand side (RHS).
secx
: Prove 1+cosxsinx+cotx=cscx
Step 1: Start with the left-hand side (LHS) and express cotx as sinxcosx.
LHS=1+cosxsinx+sinxcosx
Step 2: Combine the fractions using a common denominator.
LHS=(1+cosx)sinxsin2x+cosx(1+cosx)
Step 3: Expand the numerator.
LHS=(1+cosx)sinxsin2x+cosx+cos2x
Step 4: Use the identity sin2x+cos2x=1.
LHS=(1+cosx)sinx1+cosx
Step 5: Cancel out the common term (1+cosx).
LHS=sinx1=cscx
This matches the right-hand side (RHS).
cscx
: Prove 1−cosxsinx−cotx=cscx
Step 1: Start with the left-hand side (LHS) and express cotx as sinxcosx.
LHS=1−cosxsinx−sinxcosx
Step 2: Combine the fractions using a common denominator.
LHS=(1−cosx)sinxsin2x−cosx(1−cosx)
Step 3: Expand the numerator.
LHS=(1−cosx)sinxsin2x−cosx+cos2x
Step 4: Use the identity sin2x+cos2x=1.
LHS=(1−cosx)sinx1−cosx
Step 5: Cancel out the common term (1−cosx).
LHS=sinx1=cscx
This matches the right-hand side (RHS).
cscx
: Prove 1−sinx1+sinx=(secx+tanx)2
Step 1: Start with the right-hand side (RHS) and express secx and tanx in terms of sinx and cosx.
RHS=(cosx1+cosxsinx)2
Step 2: Combine the terms inside the parenthesis.
RHS=(cosx1+sinx)2
Step 3: Square the numerator and the denominator.
RHS=cos2x(1+sinx)2
Step 4: Use the identity cos2x=1−sin2x.
RHS=1−sin2x(1+sinx)2
Step 5: Factor the denominator using the difference of squares formula, a2−b2=(a−b)(a+b).
RHS=(1−sinx)(1+sinx)(1+sinx)2
Step 6: Cancel out one term of (1+sinx) from the numerator and denominator.
RHS=1−sinx1+sinx
This matches the left-hand side (LHS).
\frac{1+\sin x{1-\sin x}}
: Prove 1+cosx1−cosx=(cscx−cotx)2
Step 1: Start with the right-hand side (RHS) and express cscx and cotx in terms of sinx and cosx.
RHS=(sinx1−sinxcosx)2
Step 2: Combine the terms inside the parenthesis.
RHS=(sinx1−cosx)2
Step 3: Square the numerator and the denominator.
RHS=sin2x(1−cosx)2
Step 4: Use the identity sin2x=1−cos2x.
RHS=1−cos2x(1−cosx)2
Step 5: Factor the denominator using the difference of squares formula, a2−b2=(a−b)(a+b).
RHS=(1−cosx)(1+cosx)(1−cosx)2
Step 6: Cancel out one term of (1−cosx) from the numerator and denominator.
RHS=1+cosx1−cosx
This matches the left-hand side (LHS).
\frac{1-\cos x{1+\cos x}}
: Prove 1+cosxsinx=sinx1−cosx
Step 1: Start with the left-hand side (LHS) and multiply the numerator and denominator by the conjugate of the denominator, (1−cosx).
LHS=1+cosxsinx×1−cosx1−cosx
Step 2: Multiply the terms in the numerator and denominator.
LHS=12−cos2xsinx(1−cosx)
Step 3: Use the identity 1−cos2x=sin2x.
LHS=sin2xsinx(1−cosx)
Step 4: Cancel out sinx from the numerator and denominator.
LHS=sinx1−cosx
This matches the right-hand side (RHS).
\frac{1-\cos x{\sin x}}
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