4 Marks Questions

Question 514 Marks

Prove the following identities:
$\frac{(\sin\text{A}-\sin\text{B})}{(\cos\text{A}+\cos\text{B})}+\frac{(\cos\text{A}-\cos\text{B})}{(\sin\text{A}+\sin\text{B})}=0$

Answer

$\text{LHS}=\frac{(\sin\text{A}-\sin\text{B})}{(\cos\text{A}+\cos\text{B})}+\frac{(\cos\text{A}-\cos\text{B})}{(\sin\text{A}+\sin\text{B})}$
$=\frac{(\sin\text{A}+\sin\text{B})(\sin\text{A}-\sin\text{B})+(\cos\text{A}+\cos\text{B})(\cos\text{A}-\cos\text{B})}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{\sin^2\text{A}-\sin^2\text{B}+\cos^2\text{A}-\cos^2\text{B}}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{\big(\sin^2\text{A}+\cos^2\text{A}\big)-\big(\sin^2\text{A}+\cos^2\text{B}\big)}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=\frac{1-1}{(\cos\text{A}+\cos\text{B})(\sin\text{A}+\sin\text{B})}$
$=0$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$

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Question 524 Marks

Prove the following identities:
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\frac{2}{\big(\sin^2\theta-\cos^2\theta\big)}=\frac{2}{\big(2\sin^2\theta-1\big)}$

Answer

$\text{LHS}=\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}$
$=\frac{(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2}{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}$
$=\frac{\big(\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta\big)+\big(\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta\big)}{\sin^2\theta-\cos^2\theta}$
$=\frac{(1+2\sin\theta\cos\theta)+(1-2\sin\theta\cos\theta)}{\sin^2\theta-\cos^2\theta}$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$=\frac{2}{\sin^2\theta-\cos^2\theta}$
Also, $\frac{2}{\sin^2\theta-\cos^2\theta}=\frac{2}{\sin^2\theta-\big(1-\sin^2\theta\big)}$
$=\frac{2}{2\sin^2\theta-1}$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S.}$

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Question 534 Marks

Prove the following identities:
$\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}+\frac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}=\frac{2}{\big(1-2\cos^2\theta\big)}$

Answer

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Question 544 Marks

If $\sec\theta+\tan\theta=\text{p},$ prove that:
$\sin\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$

Answer

We have,
$\sec\theta+\tan\theta=\text{p}\dots(1)$
$\Rightarrow\frac{\sec\theta+\tan\theta}{1}\times\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{1}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{p}}\dots(2)$
Adding $(1)$ and $(2)$,, we get:
$2\sec\theta=\Big(\text{p}+\frac{1}{\text{p}}\Big)$
$\Rightarrow\sec\theta=\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)\dots(\text{A})$
Subtracting $(2)$ and $(1)$,, we get:
$2\tan\theta=\Big(\text{p}-\frac{1}{\text{p}}\Big)$
$\Rightarrow\tan\theta=\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)\dots(\text{B})$
Using $(A)$ and $(B)$,, we get:
$\sin\theta=\frac{\tan\theta}{\sec\theta}$
$=\frac{\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)}{\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)}$
$=\frac{\Big(\frac{\text{p}^2-1}{\text{p}}\Big)}{\Big(\frac{\text{p}^2+1}{\text{p}}\Big)}$
$\therefore\ \sin\theta=\frac{\text{p}^2-1}{\text{p}^2+1}$

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Question 554 Marks

Prove that $\frac{\sin\text{A}-2\sin^3\text{A}}{\big(2\cos^2\text{A}-\cos\text{A}\big)}=\tan\text{A}.$

Answer

$\text{LHS}=\frac{\big(\sin\text{A}-2\sin^2\text{A}\big)}{\Big(2\cos^2\text{A}-\cos\text{A}\big)}$

$=\frac{\sin\text{A}\big(1-2\sin^2\text{A}\big)}{\cos\text{A}\big(2\cos^2\text{A}-1\big)}$

$=\tan\text{A}\Bigg\{\frac{\big(\sin^2\text{A}+\cos^2\text{A}-2\sin^2\text{A}\big)}{2\cos^2\text{A}-\sin^2\text{A}-\cos^2\text{A}}\Bigg\}$ $\big[\because\sin^2\text{A}+\cos^2\text{A}=1\big]$

$=\tan\text{A}\Bigg\{\frac{\big(\cos^2\text{A}-\sin^2\text{A}\big)}{\big(\cos^2\text{A}-\sin^2\text{A}\big)}\Bigg\}$

$=\tan\text{A}$

$=\text{RHS}$

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Question 564 Marks

Prove that $\frac{1}{(sec\theta-\tan\theta)}-\frac{1}{\cos\theta}=\frac{1}{\cos\theta}-\frac{1}{(\sec\theta+\tan\theta)}.$

Answer

$\frac{1}{(\sec\theta-\tan\theta)}-\frac{1}{\cos\theta}=\frac{1}{\cos\theta}-\frac{1}{(\sec\theta+\tan\theta)}$
$\text{LHS}=\frac{1}{\sec\theta-\tan\theta}-\frac{1}{\cos\theta}$
$=\frac{(\sec\theta+\tan\theta)}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}-\sec\theta$ $($Multiplying the numerator and denominator by $\sec\theta+\tan\theta)$
$=\frac{\sec\theta+\tan\theta}{\sec^2\theta-\tan^2\theta}-\sec\theta$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\text{RHS}=\frac{1}{\cos\theta}-\frac{1}{\sec\theta+\tan\theta}$
$=\sec\theta-\frac{(\sec\theta-\tan\theta)}{\sec^2\theta-\tan^2\theta}$ $($Multiplying the numerator and denominator by $\sec\theta-\tan\theta)$
$=\sec\theta+\tan\theta-\sec\theta$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$=\tan\theta$
$\therefore\ \text{LHS}=\text{RHS}$
Hence proved.

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Maths 4 Marks Questions