Question

Prove that 1+sectheta - tan theta / 1+ sectheta + tan theta = 1-sintheta / cos theta

Answer 1

Answer:

We have to find the value of the expression:

\dfrac{1+\sec \theta -\tan \theta}{1+\sec \theta+\tan \theta}

We will rationalize our denominator:

\dfrac{(1+\sec \theta-\tan \theta)(a+\sec \theta+\tan \theta)}{(1+\sec \theta-\tan \theta)(1+\sec \theta+\tan \theta)}\\\\\\=\dfrac{(1+\sec \theta-\tan \theta)^2}{(1+\sec \theta)^2-\tan^2 \theta}\\\\\\=\dfrac{(1+\sec \theta)^2+\tan^2 \theta-2(1+\sec \theta)\tan \theta}{1+\sec^2 \theta+2\sec \theta-\tan^2 \theta}\\\\\\=\dfrac{1+\sec^2 \theta+2\sec \theta+\tan^2 \theta-2\tan \theta-2\sec \theta\tan \theta}{2+2\sec \theta}

since,

\sec^2 \theta-\tan^2 \theta=1

=\dfrac{2\sec^2 \theta+2\sec \theta-2\sec \theta\tan \theta-2\tan \theta}{2(1+\sec \theta)}\\\\\\\\=\dfrac{2\sec \theta(\sec \theta+1)-2\tan \theta(\sec \theta+1)}{2(\sec \theta+1)}\\\\\\=\dfrac{2(\sec \theta-\tan \theta)(\sec \theta+1)}{2(\sec \theta+1)}\\\\=\sec \theta-\tan \theta\\\\=\dfrac{1}{\cos \theta}-\dfrac{\sin \theta}{\cos \theta}\\\\=\dfrac{1-\sin \theta}{\cos \theta}

Hence, the value of the expression is:

\dfrac{1+\sec \theta -\tan \theta}{1+\sec \theta+\tan \theta}=\dfrac{1-\sin \theta}{\cos \theta}

Answer 2

Answer:

===>L.H.S ==> (1+ SEC Q - TAN Q) / 1 + SEC Q + TAN Q

===> WE KNOW THAT  1 = SEC²Q-TAN²Q

====> [ (SEC²Q-TAN²Q) + SEC Q - TAN Q ] / 1 + SEC Q + TAN Q

===> [ (SEC Q + TAN Q) (SEC Q - TAN Q) + SEC Q - TAN Q ] / 1+SECQ + TANQ

NOW , SEC Q - TAN Q IS COMMON

SO, SEC Q - TAN Q [ (SEC Q + TAN Q) + 1] / (1 + SECQ + TAN Q)

==> SEC Q - TAN Q [SEC Q + TAN Q + 1] / (SECQ+TANQ + 1]

==> SEC Q - TAN Q

===> 1/COSQ - SINQ/COSQ

==> (1-SINQ)/ COSQ