Question

Show that i)1+sec/sec=sin²/1-cos​

Answer 1

Solving the LHS :-

$\\\quad:\implies\quad\sf\dfrac{1+ \sec x}{\sec x}$

$\\\quad:\implies\quad\sf \dfrac{1}{\sec x} + 1$

$\\\quad:\implies\quad \boxed{\sf \cos x + 1}\bigstar$

Now Solving the RHS:-

$\\\quad:\implies\quad\sf \dfrac{\sin^{2}x }{1-\cos x}$

$\\\quad:\implies\quad\sf \dfrac{1-\cos^{2}x }{1-\cos x}$

$\\\quad:\implies\quad\sf \dfrac{(1+\cos x)(1-\cos x) }{1-\cos x}$

$\\\quad:\implies\quad \boxed{\sf \cos x + 1}\bigstar$

LHS = RHS

Hence Verified!!

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Answer 2

To Prove:

(1 + secx)/(secx) = sin²x/(1 - cosx)

Proof:

LHS:

(1 + secx)/secx

= cosx + 1

= (cosx + 1)(cosx - 1)/(1)(cosx - 1)

= (cos²x - 1)/(cosx - 1)

= sin²x/(cosx - 1) = RHS.

[As sin²x + cos²x = 1.]

H.P.

More:

• In IInd quadrant, all T-ratios are negative except sine and cosecant.
• In IIIrd quadrant, all T-ratios are negative except tangent and cotangent.
• In IVth quadrant, all T-ratios are negative except cosine and secant.