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\(A=\frac{sin2a+sin5a-sin3a}{1+cosa-2sin^22a}=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)
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\(Q=2\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+2=2\left(\frac{sina.cosa+sina}{cosa\left(cosa+1\right)}\right)^2+2\)
\(=2\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+2=2tan^2a+2=2\left(1+tan^2a\right)=\frac{2}{cos^2a}\)
\(A=\frac{sin2a+2cos4a.sina}{cos4a+cosa}=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)
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\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)
\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)
\(=1-2\left(2sinx.cosx\right)^2=1-2sin^22x=cos4x\)
\(C=\frac{cos2a+1-2cos^22a}{2sin2a.cos2a+sin2a}=\frac{\left(1-cos2a\right)\left(2cos2a+1\right)}{sin2a\left(2cos2a+1\right)}=\frac{1-cos2a}{sin2a}\)
\(=\frac{1-\left(1-2sin^2a\right)}{2sina.cosa}=\frac{2sin^2a}{2sina.cosa}=\frac{sina}{cosa}=tana\)
\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)
\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)
\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)
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a.
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cosa+sin3a}{2cos3a.cosa+cos3a}=\dfrac{sin3a\left(2cosa+1\right)}{cos3a\left(2cosa+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
b.
\(\dfrac{1+cosa}{1-cosa}.\dfrac{sin^2\dfrac{a}{2}}{cos^2\dfrac{a}{1}}-cos^2a=\dfrac{1+cosa}{1-cosa}.\dfrac{\dfrac{1-cosa}{2}}{\dfrac{1+cosa}{2}}-cos^2a\)
\(=\dfrac{1+cosa}{1-cosa}.\dfrac{1-cosa}{1+cosa}-cos^2a=1-cos^2a=sin^2a\)
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\(A=\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(B=\frac{sin^2a\left(1+tan^2a\right)}{cos^2a\left(1+cot^2a\right)}=\frac{sin^2a.\frac{1}{cos^2a}}{cos^2a.\frac{1}{sin^2a}}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(A=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)