Question

hi....can any one help in this two sums.....

... hope u guys help me also....
I'll ma u as brainliest....

Answer 1

HELLO DEAR,
(1)

cot B=12/5

AC=5cm
CB=12cm
AB=13cm

Using Pythagoras theorem
${ac }^{2} = {5}^{2} + {12}^{2} \\ = > {ac}^{2} = 169 \\ = > ac = \sqrt{169} = 13cm$

TanB=5/12
Sin B =5/13
CosB=12/13
secB=13/12

(tan²B-sin²B)=(tanB-sinB)(tanB+sinB)

=> ( 5/12-5/13) × ( 5/12 + 5/13)

=>[ (65-60)/12×13] ×[(60+65)/12×13]

=> [5/156]×[125/156]

=> 625/24336------------(1)

and

sin⁴B×sec²B

=> ( 5/13)⁴ × (13/12)²

$= > \frac{625 \times {13}^{2} }{ {13}^{2} \times {13}^{2} \times {12}^{2} } \\ = > \frac{625}{2433} ..............(2)$

from --(1) and---(2)
we get

(tan²B-sin²B) = sin⁴B×sec²B

(2)

(secA + tanA)(secB + tanB)(secC + tan C) =

(secA - tanA)(secB - tanB)(secC - tanC)

Mulitply both sides with
(secA + tanA)(secB + tanB)(secC + tan C)

we get

(secA + tanA)²(secB + tanB)²(secC + tan C)²

= (sec²A - tan²A)(sec²B - tan²B)(sec²C - tan²C)

=> (secA + tanA)²(secB + tanB)²(secC + tan C)²

=. 1×1×1=1--------(1)

similarly,

when we it with Both side by

[(secA - tanA)(secB - tanB)(secC - tanC)]

we get

(secA - tanA)²(secB - tanB)²(secC - tan C)²

= (sec²A - tan²A)(sec²B - tan²B)(sec²C - tan²C)

=> (secA - tanA)²(secB - tanB)²(secC - tan C)²
= 1×1×1=1---------(2)

form---(1) and-----(2)

(secA + tanA)(secB + tanB)(secC + tan C) =

(secA - tanA)(secB - tanB)(secC - tanC)
=+-1

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

I am so sorry I can't answer this question