Question

TRIGONOMENTRY
eliminate t between the following :-
cot t +cos t=m ,cot t - cos t=n

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Complete Question: Eliminate 't' between the following:

cot(t) + cos (t) = m

cot(t) - cos(t) = n

Given:

cot(t) + cos(t) = m

cot (t) - cos(t) = n

To Find:

The equation in terms of 'm' and 'n', after eliminating 't'.

Solution:

                           $cot (t)+cos(t)=m-Eq.(i)\\cot(t)-cos(t)=n-Eq.(ii)$

→ Adding Equations (i) and (ii):

                          $\therefore 2cot(t)=m+n\\\\\therefore cot(t)=\frac{m+n}{2} -Eq.(iii)$

→ Subtracting the Equation (ii) from Equation (i):

                           $\\\therefore 2cos(t)=m-n\\\\\therefore cos(t)=\frac{m-n}{2} -Eq.(iv)$

→ Dividing equation (iii) by equation (iv):

                           $\therefore \frac{cot(t)}{cos(t)} =\frac{(m+n)/2}{(m-n)/2} =\frac{m+n}{m-n} \\\\\therefore\frac{cos(t)}{sin(t)cos(t)} =\frac{m+n}{m-n} \\\\\therefore sin(t)=\frac{m-n}{m+n} -Eq.(v)$

→ As we know: sin²θ + cos²θ = 1

→ Therefore squaring and adding equations (iv) and (v):          

                           $\therefore sin^{2} t+cos^{2} t=1\\\\\therefore (\frac{m-n}{m+n} )^{2} +(\frac{m-n}{2} )^{2} =1$

$Hence \:the \:equation\: in\: terms\: of\: 'm'\:and\: 'n' \:is: (\frac{m-n}{m+n} )^{2} +(\frac{m-n}{2} )^{2} =1\:(Answer)$

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