Question

(1 + cot^2Q) tanQ/sec^2 = cotQ

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Prove :- [(1 + cot^2Q) tanQ]/(sec^2) = cotQ

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS,

→ [(1 + cot^2Q) tanQ]/(sec^2)

using ( 1 + cot²A) = cosec²A in Numerator we get,

→ ( cosec²A * tanA) / (sec²A)

Now, Using :-

• cosecA = (1/sinA)
• tanA = (sinA/cosA)
• secA = (1/cosA)

we get :-

→ [ (1/sin²A) * (sinA/cosA) ] / [ 1 / cos²A ]

→ (1/sinA * cosA) / (1/cos²A)

→ (1/sinA * cosA) * cos²A

→ (cosA/sinA)

→ cotA = RHS . (Proved).

Answer 2

Question:-

$(1+cot^2Q)tanQ/sec^2 = cot Q$

$\large\mathtt{SOLUTION}$

L. H. S

$\implies[( 1+cot^2Q)tanQ]/(sec^2)$

$\implies(1+cot^2A) = cosec^A$ in Numerator, We get.......

$\implies$ (cosec^A*tanA)/(sec^A)

Some useful Formulas:-

☻ cosecA = 1/sinA

☻ tanA = sinA/cosA

☻ SecA = 1/cosA

Now,

$\implies [(1/sin^2 A * ) (sinA/cosA) (1/cos^2A)]$

$\implies [(1/sinA* cosA) /(1/cos^2A)$

$\implies( cosA/sinA)$

$\implies cot A$ (proved) R. H. S

Be brainly ____♛