Prove that A -tan A - A A-

LHS #160; #160; A+tan A =cos A sin A +sin A cos A =cos #160; ^ 2 A+sin #160; ^ 2 A sin Acos A = 1 sin Acos A {∵ sin #160; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Prove that ...

Question

Prove that $cot A + tan A = sec A csc A$ .

Open in App

Solution

LHS $cot A + tan A$
$= \frac{cos A}{sin A} + \frac{sin A}{cos A}$
$= \frac{{cos}^{2} A + {sin}^{2} A}{sin A cos A}$
$= \frac{1}{sin A cos A} {∵ {sin}^{2} x + {cos}^{2} x = 1}$
$= cosec A sec A$
$= R H S$ Hence proved

Suggest Corrections

Similar questions

Q. Prove that:

tan A + cot A = sec A csc A

Q. Prove that

\frac{tan A}{1 - cot A} + \frac{cot A}{1 - tan A} = sec A csc A + 1 .

Q. Prove that:

\frac{tan A}{1 - cot A} + \frac{cos A}{1 - tan A} = 1 + tan A + cot A = 1 + sec A csc A

Prove that:

(sec A - csc A) (1 + tan A + cot A) = csc A {sec}^{2} A - cot A csc A .

$In any Δ A B C, prove that : \frac{b s e c B + c s e c C}{t a n B + t a n C} = \frac{c s e c C + a s e c A}{t a n C + t a n A} = \frac{a s e c A + b s e c B}{t a n A + t a n B}$

Join BYJU'S Learning Program

Prove that cotA+tanA=secAcscA.

LHS cotA+tanA=cosAsinA+sinAcosA=cos2A+sin2AsinAcosA=1sinAcosA{∵sin2x+cos2x=1}=cosecAsecA=RHS Hence proved

Prove that $cot A + tan A = sec A csc A$ .

LHS $cot A + tan A$
$= \frac{cos A}{sin A} + \frac{sin A}{cos A}$
$= \frac{{cos}^{2} A + {sin}^{2} A}{sin A cos A}$
$= \frac{1}{sin A cos A} {∵ {sin}^{2} x + {cos}^{2} x = 1}$
$= cosec A sec A$
$= R H S$ Hence proved