Question

17. prove that
(sin A + COSA) (tanA+cotA)=sinA-cosecA
​

Answer 1

Correct Question :-

Prove :-

$\sf ( sinA+cosA)(tanA+cotA) = secA+cosecA$

Solution :-

L.H.S = $\sf (sinA+cosA)(tanA+cotA)$

$\bullet\bf \ tanA = \dfrac{sinA}{cosA}$

$\bullet\bf \ cotA= \dfrac{cosA}{sinA}$

         $= \sf (sinA+cosA) \left( \dfrac{sinA}{cosA} +\dfrac{cosA}{sinA} \right)$

         $= \sf (sinA+cosA) \left( \dfrac{sin^2A+cos^2A}{sinAcosA} \right)$

$\bullet \bf \ sin^2A+cos^2A = 1$

         $= \sf (sinA+cosA) \times \left( \dfrac{1}{sinAcosA} \right)$

         $= \sf \dfrac{sinA}{sinAcosA} \times \dfrac{cosA}{sinAcosA}$

         $= \sf \dfrac{1}{cosA}+\dfrac{1}{sinA}$

$\bullet \bf \ secA= \dfrac{1}{cosA}$

$\bullet \bf \ cosecA= \dfrac{1}{sinA}$

         $= \sf secA+cosecA$

         = R.H.S

Hence proved.

Answer 2

Step-by-step explanation:

To prove:-

$\\\qquad\qquad\sf{:}\dashrightarrow (sinA+cosA)(tanA+cotA)=sinA+cosecA$

Proof:-

LHS:-

$\\\qquad\qquad\sf{:}\dashrightarrow (sinA+cosA)(tanA+cotA)$

$\\\qquad\qquad\sf{:}\dashrightarrow (sinA+cosA)(\dfrac {sinA}{cosA}+\dfrac {cosA}{sinA})$

$\\\qquad\qquad\sf{:}\dashrightarrow (sinA+cosA)(\dfrac{sin^2A+cos^2A}{sinA.cosA})$

$\\\qquad\qquad\sf{:}\dashrightarrow (sinA+cosA)(\dfrac {1}{sinAcosA})$

$\\\qquad\qquad\sf{:}\dashrightarrow \dfrac {sinA}{sinAcosA}\times \dfrac {cosA}{sinAcosA}$

$\\\qquad\qquad\sf{:}\dashrightarrow \dfrac {1}{cosA}+\dfrac {1}{sinA}$

$\\\qquad\qquad\sf{:}\dashrightarrow secA+cosecA$

$\\\qquad\qquad\sf{:}\dashrightarrow RHS$

$\\\therefore\sf LHS=RHS\quad {}_{(proved)}$