if tan A − cot A = 0 prove that sin A + cos A=?


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Question Number 60269 by Askash last updated on 19/May/19

$i f$ $t a n A - c o t A = 0$ $p r o v e t h a t$ $s i n A + c o s A = ?$
Commented by mr W last updated on 19/May/19

$\sin A + \cos A = 0 o r \sqrt{2} o r - \sqrt{2}$
Commented by Askash last updated on 19/May/19

$h o w - \sqrt{2}$ $i s p o s s i b l e$
Commented by mr W last updated on 19/May/19

$e . g .$ $A = π + \frac{π}{4}$ $\tan A = 1$ $\cot A = 1$ $\Rightarrow \tan A - \cot A = 0$ $\sin A = - \frac{\sqrt{2}}{2}$ $\cos A = - \frac{\sqrt{2}}{2}$ $\Rightarrow \sin A + \cos A = - \sqrt{2}$
	Answered by Kunal12588 last updated on 19/May/19

	$\frac{s i n A}{c o s A} - \frac{c o s A}{s i n A} = 0$ $\Rightarrow \frac{{s i n}^{2} A - {c o s}^{2} A}{s i n A c o s A} = 0$ $\Rightarrow (s i n A + c o s A) (s i n A - c o s A) = 0$ $\Rightarrow s i n A + c o s A = 0$
	Commented by Askash last updated on 19/May/19

	$A N O T H E R W A Y$ ${s i n}^{2} A - {c o s}^{2} A = 0$ $1 - {c o s}^{2} A - {c o s}^{2} A = 0$ $1 - 2 {c o s}^{2} A = 0$ $\frac{1}{\sqrt{2}} = c o s A$ $c o s 45 ° = c o s A$ $A = 45 °$ $s i n 45 ° + c o s 45 °$ $= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}$ $= \frac{2}{\sqrt{2}}$ $= \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$ $= \sqrt{2}$ $I S I T R I G H T O R W R O N G ? W H Y ?$
	Commented by Kunal12588 last updated on 19/May/19

	${i t}^{'} s a c t u a l l y c o r r e c t$ $b u t i t h a s t w o a n s w e r s$ $1 - 2 {c o s}^{2} A = 0$ $\Rightarrow c o s A = \pm \frac{1}{\sqrt{2}}$ $y o u s o l v e d u s i n g c a s e 1 \to c o s A = \frac{1}{\sqrt{2}}$ $n o w I a m t a k i n g c a s e 2 \to c o s A = - \frac{1}{\sqrt{2}}$ $\Rightarrow c o s A = - c o s (45 °)$ $\Rightarrow c o s A = c o s (180 ° - 45 °) = c o s (135 °)$ $\Rightarrow A = 135 ° (t a k i n g o n l y p r i c i p l e v a l u e)$ $s i n (135 °) - c o s (135 °) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$ $i n m y a n s w e r i h a d n o t t a k e n$ $(s i n A - c o s A) = 0$ $\Rightarrow s i n A + c o s A = \sqrt{2}$ $y e s - \sqrt{2} i s a l s o p o s s i b l e$ $c o s A = - c o s (45 °)$ $\Rightarrow c o s A = c o s (180 ° + 45 °) = c o s (225 °)$ $\Rightarrow A = 225 ° (i t s n o t p r i n c i p l e v a l u e)$ $s i n (225 °) + c o s (225 °) = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \sqrt{2}$
	Commented by mr W last updated on 19/May/19

	$w h a t a b o u t A = 225 ° ?$
	Commented by Kunal12588 last updated on 19/May/19

	$t h a n k y o u s i r$
	Answered by ajfour last updated on 19/May/19

	$\frac{p}{b} = \frac{b}{p} \Rightarrow p = b = \pm \frac{h}{\sqrt{2}} & p = - b = \pm \frac{h}{\sqrt{2}}$ $\sin A + \cos A = \frac{p + b}{h} = \pm \sqrt{2}, 0 .$
	Commented by Kunal12588 last updated on 19/May/19

	$g r e a t i u s e d t h i s m e t h o d t o p r o v e t h e$ $t r i g i d e n t i t i e s i n f i r s t y e a r b u t t e a c h e r s$ $a l w a y s d e d u c t m y m a r k s .$
	Commented by Askash last updated on 20/May/19

	$S a m e w i t h m e .$ $S o m e T e a c h e r s o n l y c h e c k t h e i r$ $s o l u t i o n s w h a t t h e y h a d t a u g h t .$
	Answered by malwaan last updated on 22/May/19
	$tan A−cot A=0 ⇒((sin A)/(cos A))−((cos A)/(sin A))=0 ((sin^2 A−cos^2 A)/(cos A×sin A))=0 (sin A+cos A)(sinA−cosA)=0 1: sinA+cosA=0 2: sinA−cosA=0 ⇒A=(π/4)⇒sinA+cosA=(1/(√2))+(1/(√2))=(2/(√2))=(√2) OR A=π+(π/4)=((5π)/4) ⇒sinA+cosA=−(1/(√2))−(1/(√2))=−(2/(√2))=−(√2) ∴ sinA+cosA={0 or (√2) or −(√2)}$
	$t a n A - c o t A = 0$ $\Rightarrow \frac{s i n A}{c o s A} - \frac{c o s A}{s i n A} = 0$ $\frac{{s i n}^{2} A - {c o s}^{2} A}{c o s A \times s i n A} = 0$ $(s i n A + c o s A) (s i n A - c o s A) = 0$ $1 : s i n A + c o s A = 0$ $2 : s i n A - c o s A = 0$ $\Rightarrow A = \frac{π}{4} \Rightarrow s i n A + c o s A = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ $OR A = π + \frac{π}{4} = \frac{5 π}{4}$ $\Rightarrow s i n A + c o s A = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \frac{2}{\sqrt{2}} = - \sqrt{2}$ $∴ s i n A + c o s A = {0 o r \sqrt{2} o r - \sqrt{2}}$
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