Case study (4 Marks)

Question 14 Marks

In a class test of class XI, a teacher asked to students to consider $\mathbf{A}+\mathbf{B}=\frac{\pi}{4}$, where $\mathbf{A}$ and $\mathbf{B}$ are acute angles.
Based on the above information, answer the following questions.
(i) Find the value of $(1+\tan A)(1+\tan B)$ ?
(ii) Find the value of $(\cot \mathbf{A}-1)(\cot \mathbf{B}-1)$ ?
(iii) Find the value of
$
\sin (A+B)-\cos (A+B)+\tan (A+B) .
$

Answer

(i) We have,
$
\begin{aligned}
& \mathrm{A}+\mathrm{B}=\frac{\pi}{4} \\
& \Rightarrow \tan (A+B)=\tan \frac{\pi}{4} \\
& \Rightarrow \frac{\tan A+\tan B}{1-\tan A \tan B}=1 \\
& \Rightarrow \tan A+\tan B=1-\tan A \tan B \\
& \Rightarrow \tan A+\tan B+\tan A \tan B=1 \\
&
\end{aligned}
$

On adding 1 both sides of Eq. (i), we get
$
\begin{aligned}
& & 1+\tan A+\tan B+\tan A \tan B & =2 \\
\Rightarrow & & (1+\tan A)+\tan B(1+\tan A) & =2 \\
\Rightarrow & & (1+\tan A)(1+\tan B) & =2
\end{aligned}
$

(ii) On dividing both sides of Eq. (i) by $\tan$ Atan B, we get
$
\begin{array}{ll}
& \frac{\tan A+\tan B+\tan A \tan B}{\tan A \tan B}=\frac{1}{\tan A \tan B} \\
\Rightarrow & \cot B+\cot A+1=\cot A \cot B \\
\Rightarrow & \cot A \cot B-\cot A-\cot B=1 \\
\Rightarrow & \cot A \cot B-\cot A-\cot B+1=2 \\
\Rightarrow & \cot A(\cot B-1)-(\cot B-1)=2 \\
\Rightarrow & (\cot B-1)(\cot A-1)=2
\end{array}
$

(iii) Given, $A+B=\frac{\pi}{4}$
$\begin{aligned} & \therefore \sin (A+B)=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \\ & \cos (A+B)=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\ & \text { and } \tan (A+B)=\tan \frac{\pi}{4}=1 \\ & \therefore \sin (A+B)-\cos (A+B)+\tan (A+B) \\ & =\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1=1\end{aligned}$

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Question 24 Marks

Rajiv constructs two right angled triangles in the fourth quadrant in such a way that the measure of triangle gives $\cos A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$, where $\frac{3 \pi}{2} < A$ and $B > 2 \pi$.

Based on the above information, answer the following questions.
(i) Find the value of $\cos (A+B)$
(ii) Find the value of $\sin (A-B)$
(iii) Find the value of $\tan (\mathbf{A}+\mathbf{B})$

Answer

Given, $\cos A=\frac{4}{5}$, where $\frac{3 \pi}{2}<A<2 \pi$
$
\therefore \sin A=-\sqrt{1-\cos ^2 A}=-\frac{3}{5}
$
$\left[\because\right.$ A lies in IVth quadrant] and $\cos B=\frac{12}{13}$, where $\frac{3 \pi}{2}<B<2 \pi$
$
\therefore \sin \mathrm{B}=-\sqrt{1-\cos ^2 \mathrm{~B}}=-\frac{5}{13}
$
$[\because$ B lies in IVth quadrant]

(i)
$
\begin{aligned}
\cos (A+B)= & \frac{4}{5} \times \frac{12}{13}-\left(-\frac{3}{5}\right)\left(-\frac{5}{13}\right) \\
& {[\because \cos (A+B)=\cos A \cos B-\sin A \sin B] } \\
= & \frac{48}{65}-\frac{15}{65}=\frac{33}{65}
\end{aligned}
$

(ii)
$
\begin{aligned}
\sin (A-B)= & \left(-\frac{3}{5}\right) \times \frac{12}{13}-\frac{4}{5} \times\left(-\frac{5}{13}\right) \\
& {[\because \sin (A-B)=\sin A \cos B-\cos A \sin B] } \\
= & -\frac{36}{65}+\frac{20}{65}=-\frac{16}{65}
\end{aligned}
$

(iii)
$
\begin{aligned}
\because \sin (A+B)= & \left(-\frac{3}{5}\right) \cdot\left(\frac{12}{13}\right)+\left(\frac{4}{5}\right) \cdot\left(-\frac{5}{13}\right) \\
& {[\because \sin (A+B)=\sin A \cos B+\cos A \sin B] }
\end{aligned}
$
$
\begin{aligned}
&=\frac{-36-20}{5 \times 13}=-\frac{56}{65} \\
& \therefore \tan (A+B)= \frac{-\frac{56}{65}}{\frac{33}{65}}=-\frac{56}{33} \\
& {\left[\because \tan (A+B)=\frac{\sin (A+B)}{\cos (A+B)}\right] }
\end{aligned}
$

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MATHS Case study (4 Marks)

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