Question

If cosec  + cot  = q, show that cosec  - cot  =
q
1
and hence find the values of
sin and sec .

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{cosec(\theta)+cot(\theta)=q\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)}$

Now,

We know,

$\sf{cosec^2(\theta)-cot^2(\theta)=1}$

$\sf{\implies\,\big\{cosec(\theta)-cot(\theta)\big\}\big\{cosec(\theta)+cot(\theta)\big\}=1}$

$\sf{\implies\,cosec(\theta)-cot(\theta)=\dfrac{1}{cosec(\theta)+cot(\theta)}}$

$\sf{\implies\,cosec(\theta)-cot(\theta)=\dfrac{1}{q}\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)}$

Add (1) and (2)

$\sf{2\,cosec(\theta)=q+\dfrac{1}{q}}$

$\sf{\implies\,2\,cosec(\theta)=\dfrac{q^2+1}{q}}$

$\sf{\implies\,\dfrac{2}{sin(\theta)}=\dfrac{q^2+1}{q}}$

$\sf{\implies\,\dfrac{sin(\theta)}{2}=\dfrac{q}{q^2+1}}$

$\sf{\implies\,sin(\theta)=\dfrac{2\,q}{q^2+1}}$

Subtract (2) from (1),

$\sf{2\,cot(\theta)=q-\dfrac{1}{q}}$

$\sf{\implies\dfrac{2}{tan(\theta)}=\dfrac{q^2-1}{q}}$

$\sf{\implies\dfrac{tan(\theta)}{2}=\dfrac{q}{q^2-1}}$

$\sf{\implies\,tan(\theta)=\dfrac{2\,q}{q^2-1}}$

So,

$\sf{sec(\theta)=\sqrt{tan^2(\theta)+1}}$

$\sf{\implies\,sec(\theta)=\sqrt{\left(\dfrac{2\,q}{q^2-1}\right)^2+1}}$

$\sf{\implies\,sec(\theta)=\sqrt{\dfrac{4\,q^2}{\big(q^2-1\big)^2}+1}}$

$\sf{\implies\,sec(\theta)=\sqrt{\dfrac{4\,q^2+\big(q^2-1\big)^2}{\big(q^2-1\big)^2}}}$

$\sf{\implies\,sec(\theta)=\sqrt{\dfrac{4\,q^2+q^4-2\,q^2+1}{\big(q^2-1\big)^2}}}$

$\sf{\implies\,sec(\theta)=\sqrt{\dfrac{q^4+2\,q^2+1}{\big(q^2-1\big)^2}}}$

$\sf{\implies\,sec(\theta)=\sqrt{\dfrac{\big(q^2+1\big)^2}{\big(q^2-1\big)^2}}}$

$\sf{\implies\,sec(\theta)=\dfrac{q^2+1}{q^2-1}}$