Let tan A - pp-1 and tan B - 12p - 1- if A-B-0-2 then A - B can be

Given, tan A = p(p–1) and tan B = 1(2p – 1) We know that, tan (A B) = tan A tan B1 + tan A tan B = p(p–1) nbsp; 1(2p – 1)1 +p(p–1)1(2p – 1) = (2p^2 – ...

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Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Let tan A = p...

Question

Let $tan A = \frac{p}{(p - 1)}$ and $tan B = \frac{1}{(2 p - 1)},$ if $A, B \in (0, π / 2)$ then $A - B$ can be

\frac{π}{2}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{6}

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Solution

The correct option is B $\frac{π}{4}$
Given, $tan A = \frac{p}{(p - 1)}$ and $tan B = \frac{1}{(2 p - 1)}$

We know that,
$tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B} = \frac{\frac{p}{(p - 1)} - \frac{1}{(2 p - 1)}}{1 + \frac{p}{(p - 1)} \frac{1}{(2 p - 1)}} = \frac{(2 p^{2} - 2 p + 1)}{(2 p^{2} - 2 p + 1)} = 1 = tan \frac{π}{4} \Rightarrow (A - B) = \frac{π}{4}$

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Similar questions

Q. Let

tan A = \frac{p}{(p - 1)}

and

tan B = \frac{1}{(2 p - 1)},

A, B \in (0, π / 2)

then

A - B

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Q. Let

tan A = \frac{p}{(p - 1)}

and

tan B = \frac{1}{(2 p - 1)},

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Q. Assertion :If tan A, tan B are the roots of equation

x^{2} - p x - 1 = 0

, then

{sin}^{2} (A + B) = \frac{p^{2}}{1 + p^{2}}

Reason:

{sin}^{2} (A + B) = \frac{t a n^{2} (A + B)}{1 + t a n^{2} (A + B)}

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

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π + {t a n}^{- 1} \frac{x + y}{1 - x y}

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{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

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{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

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(0, π / 4)

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(- π / 4, π / 4)

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(π / 4, π / 2)

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Let tanA=p(p–1) and tanB=1(2p–1), if A,B∈(0,π/2) then A–B can be

The correct option is B π4Given, tanA=p(p–1) and tanB=1(2p–1) We know that, tan(A−B)=tanA−tanB1+tanAtanB=p(p–1)−1(2p–1)1+p(p–1)1(2p–1)=(2p2–2p+1)(2p2–2p+1)=1=tanπ4⇒(A−B)=π4

Let $tan A = \frac{p}{(p - 1)}$ and $tan B = \frac{1}{(2 p - 1)},$ if $A, B \in (0, π / 2)$ then $A - B$ can be