Question

please solve: cosec x . cot x

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Answer 1

Answer:

ithenenn karuthunnu.

cosec x +cot x = a

=> (1/sin x)+(cos x/sin x)=a

=> 1+cos x =a.sinx

=> 1+cos x =a.√(1-cos²x )

Now, squaring both sides; we get,

1+cos²x+2.cos x = a²- a².cos²x

=> (1+a²)cos²x +2.cos x +(1-a²) = 0

Therefore,

cos x = [-2±√{4–4.(1+a²)(1-a²)}] / 2(1+a²)

=> cos x = (-1 ± a²)/(1+a²)

Answer 2

Required solution :-

Let f(x) = cosec x cot x

Using Leibnitz product rule,

$\rm f'(x)=cosec \ x(cot \ x)'+ cot \ x(cosec \ x)' \quad ...(1)$

$\rm Let \ f_1(x)=cot \ x$

$\rm f_1(x+h)=cot(x+h)$

Using the first principle,

$\rm f'_1(x)=\displaystyle \rm \lim_{h \to 0} \dfrac{f_1(x+h)-f_1(x)}{h}$

$\rm = \displaystyle \rm \lim_{h \to 0} \dfrac{cot(x+h)-cot \ x}{h}$

$\sf =\displaystyle \rm \lim_{h \to 0} \dfrac{1}{h} \bigg(\dfrac{cos(x+h)}{sin(x+h)} -\dfrac{cos \ x}{sin \ x} \bigg)$

$\rm = \displaystyle \rm \lim_{h \to 0} \dfrac{1}{h} \bigg[\dfrac{sin \ x \cos(x+h)-cos \ x \ sin(x+h)}{sin \ x \ sin(x+h)} \bigg]$

$\rm = \displaystyle \rm \lim_{h \to 0}\dfrac{1}{h} \bigg[ \dfrac{sin (x-x-h)}{sin \ x \ sin(x+h)} \bigg]$

Now we get,

$\rm =\dfrac{1}{sin \ x} \displaystyle \rm \lim_{h \to 0} \dfrac{1}{h} \bigg[ \dfrac{sin(-h)}{sin(x+h)} \bigg]$

$\rm =\dfrac{-1}{sin \ x} \bigg( \displaystyle \rm \lim_{h \to 0}\dfrac{sin \ h}{h} \bigg) \bigg(\displaystyle \rm \lim_{h \to 0}\dfrac{1}{sin(x+h)} \bigg)$

We get,

$\rm = \dfrac{-1}{sin \ x} 1 \bigg(\dfrac{1}{sin(x+0)} \bigg)$

$\rm=\dfrac{-1}{sin^2 \ x}$

$\rm =-cosec^2 \ x$

Therefore,

$\rm (cot \ x)'=-cosec^2 \ x \quad ...(2)$

Let f'₂ (x) = cosec x

Now, f₂ (x) = cosec x. Then f₂ (x + h) = cosec (x + h)

Using first principle,

$\rm =f'(x)= \displaystyle \rm \lim_{h \to 0}\frac{f_2(x+h)-f_2(x)}{h}$

$\rm = \displaystyle \rm \lim_{h \to 0} \dfrac{1}{h} (cosec(x+h)-cosec \ x)$

$\rm =\displaystyle \lim_{h \to 0} \dfrac{1}{h} \bigg[ \dfrac{1}{sin(x+h)} -\dfrac{1}{sin \ x} \bigg]$

$\rm =\displaystyle \lim_{h \to 0} \dfrac{1}{h} \bigg[ \dfrac{sin \ x -sin(x+h)}{sin \ x \ sin(x+h)} \bigg]$

Therefore,

$\rm =\dfrac{1}{sin \ x} \ \displaystyle \lim_{h \to 0} \dfrac{1}{h} \Bigg [\dfrac{2 \ cos \big(\dfrac{x+x+h}{2}\big) sin\big(\dfrac{x-x-h}{2} \big) }{sin(x+h)} \Bigg]$

$\rm =\dfrac{1}{sin \ x} \ \displaystyle \lim_{h \to 0} \dfrac{1}{h} \Bigg [\dfrac{2 \ cos \big(\dfrac{2x+h}{2}\big) sin\big(\dfrac{-h}{2} \big) }{sin(x+h)} \Bigg]$

$\rm =\dfrac{1}{sin \ x} \ \displaystyle \lim_{h \to 0} \Bigg [\dfrac{-sin \big(\dfrac{h}{2}\big) \ cos \ \big(\dfrac{2x+h}{2} \big) }{ \big( \dfrac{h}{2} \big) \ sin \ (x+h)} \Bigg]$

Now, we get

$\rm =\dfrac{-1}{sin \ x} \displaystyle \rm \lim_{h \to 0} \dfrac{sin \Big(\dfrac{h}{2}\Big) }{ \Big(\dfrac{h}{2}\Big) } \displaystyle \rm \lim_{h \to 0} \dfrac{cos \Big(\dfrac{2x+h}{2} \Big)}{sin(x+h)}$

$\rm = \dfrac{-1}{sin \ x}\ 1 \ \dfrac{cos \Big(\dfrac{2x+0}{2} \Big)}{sin(x+0)}$

$\rm =\dfrac{-1}{sin \ x} \ \dfrac{cos \ x}{sin \ x}$

$\rm =-cos \ ecx \ cot \ x$

Therefore,

$\rm (cosec \ x)'= - cos \ ecx \ cot \ x \quad ...(3)$

From (1), (2) and (3),

$\rm =f'(x)cosec \ x(-cosec^2 \ x)+cot \ x(-cosec \ x \ cot \ x)$

$\rm =-cosec^3 \ x-cot^2 \ x \ cosec \ x$