Question

Prove that:
(cosec theta-sin theta)(sec theta-cos theta)(tan theta+cot theta)=1​

Answer 1

Answer:

HAVE IT.

Step-by-step explanation: FIRST OF ALL CONVERT AL  THE RATIOS INTO SIN AND COS.

THEN U WILL HAVE TO USE THE COMMON IDENTITY [SIN$SINSQUARE THETA+COS SQUARE THETA=1$.

THEN YOU WILL SEE THAT EVERY TERM WILL GET CANCELLED.

AND AT LAST YOU WILL BE LEFT WITH UNITY!!!

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

$1 - {cos}^{2} \theta \: = {sin}^{2} \theta \:$

2.

$1 - {sin}^{2} \theta \: = {cos}^{2} \theta \:$

3.

$1 + {tan}^{2} \theta \: = {sec}^{2} \theta \:$

CALCULATION

$(cosec \theta-sin \theta)(sec \theta-cos \theta)(tan \theta+cot \theta)$

$\displaystyle \: = ( \frac{1}{sin \theta}-sin \theta)(\frac{1}{cos \theta}-cos \theta)(tan \theta+\frac{1}{tan \theta})$

$\displaystyle \: = ( \frac{1-{sin}^{2} \theta}{sin \theta})( \frac{1-{cos}^{2} \theta}{cos \theta})( \frac{1 + {tan}^{2} \theta}{tan \theta})$

$\displaystyle \: = ( \frac{{cos}^{2} \theta}{sin \theta})( \frac{{sin}^{2} \theta}{cos \theta})( \frac{ {sec}^{2} \theta}{tan \theta})$

$\displaystyle \: = ( \frac{{cos}^{2} \theta}{sin \theta})( \frac{{sin}^{2} \theta}{cos \theta}) \times \frac{1}{{cos}^{2} \theta} \times \frac{cos \theta}{sin \theta}$

$= 1$