Question

☆☆☆☆Need help☆☆☆☆


If A+B = 45°, then (1+tan A)(1+tanB) should be equal to ___________.

Explain your answer.

Answer 1

Hey !!!

A + B = 45° (given )

➡ As, (1 + tanA ) ( 1 + tanB)

= (1 + 1/cotA ) (1 + 1 /cotB )

= (cotA + 1 /cotA ) ( 1 +1 /cotB)

= (cotA + 1 /cotA ) ( 1+1/cot ( 45° - A)
(•°• A + B = 45°)

= (cotA + 1 /cotA ){ 1 +( cotA - cot45° /cot45° ×cotA + 1 )}

=•°•cot(A+ B) = cotA ×cotB + 1 /cotB - cotA
using this identity

= cotA + 1 /cotA{ ( 1 + (cotA -1 /cotA +1}

= cotA + 1 /cotA {( cotA +1 + cotA -1 )} /cotA +1

= (cotA + 1 /cotA )(cotA+ 1 + cotA -1 )/cotA + 1

= ( cotA + 1 /cotA) ( 2cotA/cotA +1 )

= then( cotA+ 1 ) cancelled . and cotA is also cancelled from numerator to denominator

remaining 2

so, the answer is 2

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Hope it helps you !!!

@Rajukumar111

Answer 2

HELLO DEAR,

we know that:-

$tan( \alpha + \beta ) = \frac{tan \alpha + tan \beta }{1 - tan \alpha tan \beta }$

(A + B) = 45°

=> tan(A + B) = tan45°

$\frac{tan \alpha + tan \beta }{1 - tan \alpha tan \beta } = tan45 \\ \\ = > \frac{tan \alpha + tan \beta }{1 - tan \alpha tan \beta } = 1 \\ \\ = > tan \alpha + tan \beta = 1 - tan \alpha \: tan \beta \\ \\ = > tan \alpha + tan \beta + tan \alpha \: tan \beta = 1 \\ \\ = > 1 + tan \alpha + tan \beta + tan \alpha \: tan \beta = 1 + 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [adding \: both \: side \: (1)] \\ \\ = > 1(1 + tana ) \: + tan \beta (1 + tan \alpha ) = 2 \\ \\ = > (1 + tan \alpha )(1 + tan \beta ) = 2$

I HOPE ITS HELP YOU DEAR,
THANKS