1 - 1tan2A 1 - 1cot2A - 1sin2A sin4A

Given, (1+1tan^2A)(1+1cot^2A) =(1+cot^2A)(1+tan^2A) =(cosec^2A)(sec^2A) =1(sin^2A)(cos^2A) =1(sin^2A)(1 sin^2A) =1sin^2A sin^4A=RH ...

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

1 + 1tan2A ...

Question

$(1 + \frac{1}{t a n^{2} A})$ $(1 + \frac{1}{c o t^{2} A})$ = $\frac{1}{s i n^{2} A s i n^{4} A}$

Open in App

Solution

given, $(1 + \frac{1}{t a n^{2} A}) (1 + \frac{1}{c o t^{2} A})$

$= (1 + c o t^{2} A) (1 + t a n^{2} A)$

$= (c o s e c^{2} A) (s e c^{2} A)$

$= \frac{1}{(s i n^{2} A) (c o s^{2} A)}$

$= \frac{1}{(s i n^{2} A) (1 - s i n^{2} A)}$

$= \frac{1}{s i n^{2} A s i n^{4} A} = R H S$

Suggest Corrections

Similar questions

Q. Prove that

\frac{1 + t a n^{2} A}{1 + c o t^{2} A}

t a n^{2} A .

Q. show that

\frac{1 - {t a n}^{2} A}{{c o t}^{2} A - 1} = {t a n}^{2} A

$\frac{1 + t a n^{2} A}{1 + c o t^{2} A}$ =

Q. Prove that:

(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = {(\frac{1 - t a n A}{1 - c o t A})}^{2} = t a n^{2} A

Question 5 (x)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{1 - c o t A})^{2} = t a n^{2} A$

(1+1tan2A) (1+1cot2A) = 1sin2Asin4A

given, (1+1tan2A)(1+1cot2A)=(1+cot2A)(1+tan2A)=(cosec2A)(sec2A)=1(sin2A)(cos2A)=1(sin2A)(1−sin2A)=1sin2Asin4A=RHS

$(1 + \frac{1}{t a n^{2} A})$ $(1 + \frac{1}{c o t^{2} A})$ = $\frac{1}{s i n^{2} A s i n^{4} A}$

given, $(1 + \frac{1}{t a n^{2} A}) (1 + \frac{1}{c o t^{2} A})$

$= (1 + c o t^{2} A) (1 + t a n^{2} A)$

$= (c o s e c^{2} A) (s e c^{2} A)$

$= \frac{1}{(s i n^{2} A) (c o s^{2} A)}$

$= \frac{1}{(s i n^{2} A) (1 - s i n^{2} A)}$

$= \frac{1}{s i n^{2} A s i n^{4} A} = R H S$