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Please answer 14 and 15 question of the image below

Please answer 14 and 15 question of the image below

Question Image
Grade:10

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
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
Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
7 years ago
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.

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