Question

solve this 9 sums please​

Answer 1

Answer:

sorry I don't know the answer

Answer 2

Question:-

Prove that

$\frac{ {tan}^{3}a - 1 }{tan \: a - 1} = {sec}^{2}a + tan \: a$

Solution:-

We have,

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )$

So,

${tan}^{3} a - {1}^{3} = (tan \: a \: - 1)( {tan}^{2} a + 1 + tan \: a \times 1)$

$\frac{(tan \: a - 1)( {tan}^{2}a + 1 + tan \: a) }{(tan \: a - 1)}$

Numerator tan a-1 and denominator tan a-1 cancel

${tan}^{2} a + 1 + tan \: a$

We have,

${tan}^{2} a + 1 = {sec}^{2} a$

So, It becomes

${sec}^{2} a + tan \: a$

LHS = RHS

Hence, Proved