Question

(cosec A-sin A) (sec A-cos A) (tan A+cot A)=1

Answer 1

I solve this question

=(1/sinA-sinA)(1/cosA-cosA)(sinA/cosA+cosA/sinA)
take l.c.m.
you have
1-sin^2A/sinA×1-cos^2A/cosA×sin^2A+cos^2A/cosAsinA
then you know that sin^2A+cos^2A=1
then the last step is that


cos^2A/sinA×sin^2A/cosA×1/cosAsinA

cancelling we have cotA×tanA then this the Identity =1

Answer 2

Question:

(cosec A - sin A) (sec A - cos A) (tan A + cot A) = 1

Solution:

$= (\dfrac{1}{sinA} - sinA) (\dfrac{1}{cosA} - cosA) (\dfrac{sinA}{cosA} + \dfrac{cosA}{sinA})$

$= (\dfrac{1-sin^{2}A}{sinA})(\dfrac{1-cos^{2}A}{cosA})(\dfrac{sin^{2}A + cos^{2}A}{sinA\:cosA}$

$= (\dfrac{cos^{2}A}{sinA})(\dfrac{sin^{2}A}{cosA})(\dfrac{1}{sinA\:cosA})$

$= \dfrac{cosA \: sinA}{sinA \: cosA} = 1$

Hence, proved.

☆Identities used☆

• $cosec\theta = \dfrac{1}{sin\theta}$

• $sec\theta = \dfrac{1}{cos\theta}$

• $sin^{2}\theta + cos^{2}\theta = 1$