Let A - B - C are three angles of a triangle such that A -4 and tan B tan C - p- Then the minimum positive value of - p - is where - is the greatest integer function

A+B+C = π ⇒ B+C = 3π4 So, 0 B, C 3π4 Now, tan B tan C = p ⇒sin B sin Ccos B cos C = p ⇒cos B cos C sin B sin Ccos B cos C +sin B sin C = 1 p 1 nbsp; + p ...

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Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Let A , B , C...

Question

Let $A, B, C$ are three angles of a triangle such that $A = \frac{π}{4}$ and $tan B tan C = p$ . Then the minimum positive value of $[p]$ is
(where $[.]$ is the greatest integer function)

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Solution

$A + B + C = π \Rightarrow B + C = \frac{3 π}{4}$
So, $0 < B, C < \frac{3 π}{4}$

Now, $tan B tan C = p$
$\Rightarrow \frac{sin B sin C}{cos B cos C} = p \Rightarrow \frac{cos B cos C - sin B sin C}{cos B cos C + sin B sin C} = \frac{1 - p}{1 + p} \Rightarrow \frac{cos (B + C)}{cos (B - C)} = \frac{1 - p}{1 + p} \Rightarrow cos (B - C) = - \frac{1 + p}{\sqrt{2} (1 - p)}$

We know that,
$0 \leq B - C < \frac{3 π}{4} \Rightarrow - \frac{1}{\sqrt{2}} < cos (B - C) \leq 1 \Rightarrow - \frac{1}{\sqrt{2}} < - \frac{1 + p}{\sqrt{2} (1 - p)} \leq 1 \Rightarrow - \sqrt{2} \leq \frac{1 + p}{(1 - p)} < 1$

$\frac{1 + p}{(1 - p)} \geq - \sqrt{2} \Rightarrow \frac{p + 1}{p - 1} - \sqrt{2} \leq 0 \Rightarrow \frac{p - (\sqrt{2} + 1)^{2}}{p - 1} \geq 0 \Rightarrow p < 1 or p \geq (\sqrt{2} + 1)^{2} \dots (1)$

$\frac{1 + p}{(1 - p)} < 1 \Rightarrow \frac{2 p}{p - 1} > 0 \Rightarrow p < 0 or p > 1 \dots (2)$

From $(1)$ and $(2)$ ,
$p < 0 or p \geq (\sqrt{2} + 1)^{2}$
Hence, the minimum positive value of $[p]$ is $5$ .

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Let A,B,C are three angles of a triangle such that A=π4 and tanBtanC=p. Then the minimum positive value of [p] is (where [.] is the greatest integer function)

Let $A, B, C$ are three angles of a triangle such that $A = \frac{π}{4}$ and $tan B tan C = p$ . Then the minimum positive value of $[p]$ is
(where $[.]$ is the greatest integer function)