Question

1.)Tan (A+B)=π/4 prove that

(1+tanA)(1+TanB)=2


2.)sinπ/6+cos^2π/3-tan^2π/4=1/2


3.)tan15°+tan30°+tan15°tan30°=1

Answer 1

See The attachment I think it will help you ^_^

✴In question first and third ,we use the identity

Tan(A+B)= TanA+Tan B/1-tanAtanB

✴In question Second

we, know that π/6=30°

cos^2π/3=60°


or , tan^2 π/4=45°

THANKS ❤^_^


#RÔYÂL CHÔRÎ ♥

Answer 2

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$\large\red{Question:}$

If (A+B)=$\frac{\pi}{4}$ prove that (1+tanA)(1+TanB)=2

$\huge\red{Answer:}$

$tan (A+B)= \tan\: \frac{\pi}{4}=1$

Using the Identity,
$tan\: (A+B) \frac{tan\: A+tan\: B}{1-tan\: A.tan\: B}$

$1 = \frac{tan\: A+tan\: B}{1-tan\: A.tan\: B}$

$tan\: A+tan\: B = 1-tan\: A.tan\: B$

$tan\: A+tan\: B + tan\: A.tan\: B = 1$

$tan\: A+tan\: B(1 + tan\: A)= 1$

Add 1 on both sides
$(1+tan\: A)+tan\: B(1 + tan\: A)= 2$

$(1+tan\: A)(1+tan\: B) = 2$

$\large\red{Question:}$
$Sin^{2}\: \frac{\pi}{6}+ Cos^{2}\: \frac{\pi}{3}-tan^{2}\: \frac{\pi}{4}$

$\huge\red{Answer:}$

We have,
$Sin\: \frac{\pi}{6}=\frac{1}{2}$

$Cos\: \frac{\pi}{3} = \frac{1}{2}$

$tan\: \frac{\pi}{4} = 1$

=> $(\frac{1}{2})^{2}+(\frac{1}{2})^{2} -1^{2}$

= $\frac{1}{4}+ \frac{1}{4} - 1$

= $\frac{2}{4} - 1$

= $\frac{1}{2} - 1$

= $-\frac{1}{2}$

$\large\red{Question:}$
tan 15°+tan 30°+tan 15°tan30°=1

We have,
tan 45° = 1
=> tan (30+15) = 1

Using the Identity,
$tan\: (A+B) = \frac{tan\: A+tan\: B}{1-tan\: A.tan\: B}$

$tan\: (30+15)= \frac{tan\: 30+tan\: 15}{1-tan\: 30.tan\: 15}$

=> $1 = \frac{tan\: 30+tan\: 15}{1-tan\: 30.tan\: 15}$

=> $tan\: 30+tan\: 15 = 1-tan\: 30.tan\: 15$

=> $tan\: 30+tan\: 15 + tan\: 30.tan\: 15 = 1$

✓✓✓

$\huge\boxed{\texttt{\fcolorbox{red}{aqua}{Keep\: it\: Mello}}}$

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