If cos-1 p - cos-1q - cos-1 r - - thenp2 - q2 - r2 - 2pqr-

Trick: According to given condition, we put p = q = r = 1/2. p^2 + q^2 + r^2 + 2pqr = ( 1/2 )^2 + ( 1/2 )^2 + ( 1/2 )^2 + 2.1/2. 1/2.1/2. = 1/4 + 1/4 + 1/ ...

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Standard XIII

Mathematics

Properties Derived from Trigonometric Identities

If cos-1 p + ...

Question

If ${cos}^{- 1} p + {cos}^{- 1} q + {cos}^{- 1} r = π$ then
$p^{2} + q^{2} + r^{2} + 2 p q r =$

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Solution

The correct option is B 1
Trick: According to given condition, we put
$p = q = r = \frac{1}{2} . p^{2} + q^{2} + r^{2} + 2 p q r = {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} + 2. \frac{1}{2} . \frac{1}{2} . \frac{1}{2} . = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{2}{8} = 1.$

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Similar questions

State whether the statement is true/false.

{cos}^{- 1} p + {cos}^{- 1} q + {cos}^{- 1} r = π

, then

p^{2} + q^{2} + r^{2} + 2 p q r = 1

If $c o s^{- 1} p + c o s^{- 1} q + c o s^{- 1} r = π$ then $p^{2} + q^{2} + r^{2} =$

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}

x y < 1

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

x y > 1

(a) If

{c o s}^{- 1} p + {c o s}^{- 1} q + {c o s}^{- 1} r = π

, then prove that

p^{2} + q^{2} + r^{2} + 2 p q r = 1

(b) If

{s i n}^{- 1} x + {s i n}^{- 1} y + {s i n}^{- 1} z = π

, then prove that

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

{t a n}^{- 1} x + {t a n}^{- 1} y + {t a n}^{- 1} z = π

π / 2

show that

x + y + z = x y z

x y + y z + z x = 1

Q. If

c o s^{- 1} \frac{p}{a} + c o s^{- 1} \frac{q}{b} = α

, then

\frac{p^{2}}{a^{2}} - \frac{2 p q}{a b} c o s α + \frac{q^{2}}{b^{2}}

is equal to

Q. If

{cos}^{- 1} \frac{p}{a} + {cos}^{- 1} \frac{q}{b} = α

, then prove that

\frac{p^{2}}{a^{2}} - \frac{2 p q}{a b} cos α + \frac{q^{2}}{b^{2}} = {sin}^{2} α

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If cos−1p+cos−1q+cos−1r=π then p2+q2+r2+2pqr=

The correct option is B 1Trick: According to given condition, we put p=q=r=12.p2+q2+r2+2pqr=(12)2+(12)2+(12)2+2.12.12.12.=14+14+14+28=1.

If ${cos}^{- 1} p + {cos}^{- 1} q + {cos}^{- 1} r = π$ then
$p^{2} + q^{2} + r^{2} + 2 p q r =$