Show that 1-cosecA-cotA - 1-sinA - 1-sinA-1-cosecA-cotA

Step 1: Rationalizing the denominator of 1/(cosecA cotA), we get 1/(cosecA cotA) =1/(cosecA cotA)×(cosecA+cotA)/(cosecA+cotA) ...

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Byju's Answer

Standard XII

Mathematics

Sin(A+B)Sin(A-B)

Show that 1/c...

Question

Show that $\frac{1}{(cosecA - cotA)} - \frac{1}{sinA} = \frac{1}{sinA} - \frac{1}{(cosecA + cotA)}$

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Solution

Step $1$ : Rationalizing the denominator of $\frac{1}{(cosecA - cotA)}$ , we get
$\frac{1}{(cosecA - cotA)}$ $= \frac{1}{(cosecA - cotA)} \times \frac{(cosecA + cotA)}{(cosecA + cotA)}$
$= \frac{(cosecA + cotA)}{({cosec}^{2} A - \cot^{2} A)}$ [Identity used : $a^{2} - b^{2} = (a + b) (a - b)$ ]
$= \frac{(cosecA + cotA)}{1}$ [Identity used : ${cosec}^{2} A - \cot^{2} A = 1$ ]
$= (cosecA + cotA)$
Step $2$ : Similarly, rationalizing the denominator of $\frac{1}{(cosecA + cotA)}$ , we get
$\frac{1}{(cosecA - cotA)}$ $= \frac{1}{(cosecA + cotA)} \times \frac{(cosecA - cotA)}{(cosecA - cotA)}$
$= \frac{(cosecA - cotA)}{({cosec}^{2} A - \cot^{2} A)}$ [Identity used : $a^{2} - b^{2} = (a + b) (a - b)$ ]
$= \frac{(cosecA - cotA)}{1}$ [Identity used : ${cosec}^{2} A - \cot^{2} A = 1$ ]
$= (cosecA - cotA)$
Step $3$ : In the trigonometric equation, $\frac{1}{(cosecA - cotA)} - \frac{1}{sinA} = \frac{1}{sinA} - \frac{1}{(cosecA + cotA)}$ , we have
$L . H . S = \frac{1}{(cosecA - cotA)} - \frac{1}{sinA}$ $= cosecA + cotA - \frac{1}{sinA}$
$= cosecA + cotA - cosecA$ [Identity used : $\frac{1}{sinA} = cosecA$ ]
$= cotA$
$R . H . S = \frac{1}{sinA} - \frac{1}{(cosecA - cotA)}$ $= \frac{1}{sinA} - (cosecA - cotA)$
$= cosecA - cosecA + cotA$
$= cotA$
From above it is clear that $L . H . S = R . H . S$
hence it is proved that $\frac{1}{(cosecA - cotA)} - \frac{1}{sinA} = \frac{1}{sinA} - \frac{1}{(cosecA + cotA)}$

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Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

, where

Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Q. Among

\frac{1}{12}

\frac{1}{15}

and

\frac{1}{17}

, is the greatest fraction

Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

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