Trigonometry> Please answer 14 and 15 question of the i...

Please answer 14 and 15 question of the image below

Viraj C Bukitagar , 11 Years ago

Grade 10

2 Answers

Jitender Singh

Hello student,
Please find answer to your question
14.
$\frac{tanA}{1-cotA} + \frac{cotA}{1-tanA} = secA.cosecA + 1$
$LHS = \frac{tanA}{1-cotA} + \frac{cotA}{1-tanA}$
$LHS = \frac{tanA}{1-\frac{1}{tanA}} + \frac{cotA}{1-tanA}$
$LHS = \frac{tan^{2}A}{tanA-1} + \frac{-cotA}{tanA-1}$
$LHS = \frac{tan^{2}A-cotA}{tanA-1}$
$LHS = \frac{tan^{3}A-1}{tanA(tanA-1)}$
$LHS = \frac{(tanA-1)(tan^{2}A + 1 + tanA)}{tanA(tanA-1)}$
$LHS = \frac{(tan^{2}A + 1 + tanA)}{tanA}$
$LHS = tanA + cotA + 1$
$LHS = \frac{sinA}{cosA} + \frac{cosA}{sinA} + 1$
$LHS = \frac{sin^{2}A + cos^{2}A}{sinAcosA} + 1$
$LHS = \frac{1}{sinAcosA} + 1$
$LHS = secA.cosecA + 1$

15.
$\frac{cosA}{1-tanA} + \frac{sinA}{1-cotA} = sinA + cosA$
$LHS = \frac{cosA}{1-tanA} + \frac{sinA}{1-cotA}$
$LHS = \frac{cosA}{1-tanA} + \frac{sinA}{1-\frac{1}{tanA}}$
$LHS = \frac{-cosA}{tanA-1} + \frac{tanA.sinA}{tanA-1}$
$LHS = \frac{sin^{2}A-cos^{2}A}{cosA(tanA-1)}$
$LHS = \frac{(sinA-cosA)(sinA + cosA)}{(sinA-cosA)}$
$LHS = sinA + cosA$

Last Activity: 11 Years ago

Sher Mohammad

please upload a different image , the questions in the image are not clear, i can’t help you untill you post the questions clearly, please post the questions clearly.

Last Activity: 11 Years ago