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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions4 Marks
Question
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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