If $t a n^{2} A = 2 t a n^{2} B + 1$ then the value of $c o s 2 A + s i n^{2} B$ .

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Solution

We are given $t a n^{2} A = 2 t a n^{2} B + 1$

We want to find $c o s 2 A + s i n^{2} B$

We are given the value of $t a n^{2} A$ .So, write the value of cos2A in terms of tanA

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

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

Substituting the value of $t a n^{2} A$

= $\frac{1 - (2 t a n^{2} B + 1)}{1 + 2 t a n^{2} B + 1} + s i n^{2} B$

= $\frac{- 2 t a n^{2} B}{2 (1 + t a n^{2} B} + s i n^{2} B$

= $\frac{- 2 t a n^{2} B}{s e c^{2} B} + s i n^{2} B$

= $- s i n^{2} B + s i n^{2} B$

= 0

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If tan2A=2tan2B+1 then the value of cos2A+sin2B. __

If $t a n^{2} A = 2 t a n^{2} B + 1$ then the value of $c o s 2 A + s i n^{2} B$ .

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