[5 marks sum]

Question 15 Marks

Prove the following identity :
$
\frac{\cot ^2 \theta(\sec \theta-1)}{(1+\sin \theta)}=\sec ^2 \theta\left(\frac{1-\sin \theta}{1+\sec \theta}\right)
$

Answer

$
\begin{aligned}
& \text { LHS }=\frac{\cot ^2 \theta(\sec \theta-1)}{(1+\sin \theta)} \\
& =\frac{\cot ^2 \theta(\sec \theta-1)(1-\sin \theta)(\sec \theta+1)}{(1+\sin \theta)(1-\sin \theta)(\sec \theta+1)} \\
& =\frac{\cot ^2 \theta(\sec \theta-1)(\sec \theta+1)(1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)(\sec \theta+1)} \\
& =\frac{\cot ^2 \theta\left(\sec ^2 \theta-1\right)(1-\sin \theta)}{\left(1-\sin ^2 \theta\right)(1+\sec \theta)} \\
& =\frac{\cot ^2 \theta\left(\tan ^2 \theta\right)(1-\sin \theta)}{\left(\cos ^2 \theta\right)(1+\sec \theta)} \quad\left(\because \tan ^2 \theta=\sec ^2 \theta-1,1-\sin ^2 \theta=\cos ^2 \theta\right) \\
& =\frac{\left(\cot ^\theta \tan ^2\right)}{\left(\cos ^2 \theta\right)(1+\sin \theta)} \\
& =\frac{1(1-\sin \theta)}{\left(\cos ^2 \theta\right)(1+\sec \theta)} \quad(\because \cot \theta \tan \theta=1) \\
& =\sec ^2 \theta\left(\frac{1-\sin \theta}{1+\sec \theta}\right)
\end{aligned}
$

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Question 25 Marks

Prove the following identity :
$
\left[\frac{1}{\left(\sec ^2 \theta-\cos ^2 \theta\right)}+\frac{1}{\left(\operatorname{cosec} 2-\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)=\frac{1-\sin ^2 \theta \cos ^2 \theta}{2+\sin ^2 \theta \cos ^2 \theta}
$

Answer

$
\begin{aligned}
& \text { LHS }=\left[\frac{1}{\left(\sec ^2 \theta-\cos ^2 \theta\right)}+\frac{1}{\left(\operatorname{cosec} 2-\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right) \\
& =\left[\frac{1}{\left(\frac{1}{\cos ^2 \theta}-\cos ^2 \theta\right)}+\frac{1}{\left(\frac{1}{\sin ^2 \theta}-\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)
\end{aligned}
$
$
\begin{aligned}
& =\left[\frac{1}{\left(\frac{1-\cos ^4 \theta}{\cos ^2 \theta}\right)}+\frac{1}{\left(\frac{1-\sin ^4 \theta}{\sin ^2 \theta}\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right) \\
& \left.=\left[\frac{\cos ^2 \theta}{1-\cos ^4 \theta}+\frac{\sin ^2 \theta}{1-\sin ^4 \theta}\right)\right]\left(\sin ^2 \theta \cos ^2 \theta\right)
\end{aligned}
$
$
\begin{aligned}
& \left.=\left[\frac{\cos ^2 \theta-\cos ^2 \theta \sin ^2 \theta+\sin ^2 \theta-\sin ^2 \theta \cos ^4 \theta}{\left(1-\cos ^4 \theta\right)\left(1-\sin ^4 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)\right] \\
& =\left[\frac{\cos ^2 \theta+\sin ^2 \theta-\cos ^2 \theta \sin ^2 \theta\left(\cos ^2 \theta+\sin ^2 \theta\right)}{\left(1-\cos ^2 \theta\right)\left(1+\cos ^2 \theta\right)\left(1-\sin ^2 \theta\right)\left(1+\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right) \\
& =\left[\frac{1-\cos ^2 \theta \sin ^2 \theta}{\sin ^2 \theta\left(1+\cos ^2 \theta\right) \cos ^2 \theta\left(1+\sin ^2 \theta\right)}\right]\left(\sin ^2 \theta \cos ^2 \theta\right)
\end{aligned}
$
$
\begin{aligned}
& \left(\because \cos ^2 \theta+\sin ^2 \theta=1,\left(1-\cos ^2 \theta\right)=\sin ^2 \theta,\left(1-\sin ^2 \theta\right)=\cos ^2 \theta\right) \\
& =\frac{1-\cos ^2 \theta \sin ^2 \theta}{\left(1+\cos ^2 \theta\right)\left(1+\sin ^2 \theta\right)}=\frac{1-\cos ^2 \theta \sin ^2 \theta}{1+\sin ^2 \theta+\cos ^2 \theta+\sin ^2 \theta \cos ^2 \theta} \\
& =\frac{1-\cos ^2 \theta \sin ^2 \theta}{1+1+\sin ^2 \theta \cos ^2 \theta}=\frac{1-\sin ^2 \theta \cos ^2 \theta}{2+\sin ^2 \theta \cos ^2 \theta}
\end{aligned}
$

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Question 35 Marks

Prove the following identity:
$
\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta}=2
$

Answer

$
\begin{aligned}
& \text { LHS }=\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta} \\
& =\frac{\left(\cos ^3 \theta+\sin ^3 \theta\right)(\cos \theta-\sin \theta)+\left(\cos ^3 \theta-\sin ^3 \theta\right)(\cos \theta+\sin \theta)}{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)} \\
& =\frac{\cos ^4 \theta-\cos ^3 \theta \sin \theta+\sin ^3 \theta \cos \theta-\sin ^4 \theta+\cos ^4 \theta+\cos ^3 \theta \sin \theta-\sin ^3 \theta \cos \theta-\sin ^4 \theta}{\cos ^2 \theta-\sin ^2 \theta} \\
& =\frac{2 \cos ^4 \theta-2 \sin ^4 \theta}{\cos ^2 \theta-\sin ^2 \theta}=\frac{2\left(\cos ^4 \theta-\sin ^4 \theta\right)}{\cos ^2 \theta-\sin ^2 \theta} \\
& =\frac{2\left(\cos ^2 \theta+\sin ^2 \theta\right)\left(\cos ^2 \theta-\sin ^2 \theta\right)}{\left(\cos ^2 \theta-\sin ^2 \theta\right)}=2\left(\cos ^2 \theta+\sin ^2 \theta\right) \\
& =2
\end{aligned}
$
OR
$
\begin{aligned}
& \text { LHS }=\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta} \\
& =\frac{(\cos \theta+\sin \theta)\left(\cos ^2 \theta+\sin ^2 \theta-\cos \theta \sin \theta\right)}{\cos \theta+\sin \theta}+\frac{(\cos \theta-\sin \theta)\left(\cos ^2 \theta+\sin ^2 \theta+\cos \theta \sin \theta\right)}{(\cos \theta-\sin \theta)} \\
& \left(\because a^3 \pm b^3=(a \pm b)\left(a^2+b^2 \pm a b\right)\right) \\
& =\left(\cos ^2 \theta+\sin ^2 \theta-\cos \theta \sin \theta\right)+\left(\cos ^2 \theta+\sin ^2 \theta+\cos \theta \sin \theta\right) \\
& =1-\cos \theta \sin \theta+1+\cos \theta \sin \theta \quad\left(\because \cos ^2 \theta+\sin ^2 \theta=1\right) \\
& =2
\end{aligned}
$

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Question 45 Marks

Prove the following identity:
$
\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin A}=2
$

Answer

$
\begin{aligned}
& \text { LHS }=\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin A} \\
& =\frac{\left(\cos ^3 A+\sin ^3 A\right)(\cos A-\sin A)+\left(\cos ^3 A-\sin ^3 A\right)(\cos A+\sin A)}{\cos ^2 A-\sin ^2 A}
\end{aligned}
$
$
\begin{aligned}
& =\frac{\cos ^4 A-\cos ^3 A \sin A+\sin ^3 A \cos A-\sin ^4 A+\cos ^4 A+\cos ^3 A \sin A-\sin ^3 A \cos A=\sin ^4 A}{\cos ^2 A-\sin ^2 A} \\
& =\frac{2\left(\cos ^4 A-\sin ^4 A\right)}{\cos ^2 A-\sin ^2 A}=\frac{2\left(\cos ^2 A+\sin ^2 A\right)\left(\cos ^2 A-\sin ^2 A\right)}{\left(\cos ^2 A-\sin ^2 A\right)}=2\left(\cos ^2 A+\sin ^2 A\right) \\
& =2\left(\because \cos ^2 A+\sin ^2 A=1\right)
\end{aligned}
$
OR
$
\begin{aligned}
& =\frac{\cos ^3 A+\sin ^3 A}{\cos A+\sin A}+\frac{\cos ^3 A-\sin ^3 A}{\cos A-\sin A} \\
& = \frac{(\cos A+\sin A)\left(\cos ^2 A+\sin ^2 A-\cos A \sin A\right)}{(\cos A+\sin A)}+\frac{(\cos A-\sin A)\left(\cos ^2 A+\sin ^2 A+\cos A \sin A\right)}{(\cos A-\sin A)} \\
& \left(\because \mathrm{a}^3 \pm \mathrm{b}^3=(\mathrm{a} \pm \mathrm{b})\left(\mathrm{a}^2+\mathrm{b}^2 \pm \mathrm{ab}\right)\right) \\
& =\left(\cos ^2 A+\sin ^2 A-\cos A \sin A\right)+\left(\cos ^2 A+\sin ^2 A+\cos A \sin A\right) \\
& =1-\cos A \sin A+1+\cos A \sin A\left(\because \cos ^2 A+\sin ^2 A=1\right) \\
& =2
\end{aligned}
$

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Mathematics [5 marks sum]

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