Question

Trigonometry:
If A is an acute angle in a right
ABC, right angled at B, then the value of
Sin A + cos A is
(A) equal to one
C) less than one
(B) greater than one
(D) equal to two
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A is an acute angle in right angled triangle ABC, right angled at B.

Now, we know that

$\rm \: sinA = \dfrac{opposite \: side}{hypotenuse}$

$\rm\implies \:sinA = \dfrac{BC}{AC}$

Also, we know that

$\rm \: cosA = \dfrac{adjacent \: side}{hypotenuse}$

$\rm\implies \:cosA \: = \: \dfrac{AB}{AC}$

Now, Consider

$\rm \: sinA + cosA$

$\rm \: = \: \dfrac{BC}{AC} + \dfrac{AB}{AC}$

$\rm \: = \: \dfrac{BC + AB}{AC}$

We know,

In triangle, sum of any two sides is greater than third side.

$\rm \: > \: \dfrac{AC}{AC}$

$\rm \: > \: 1$

Hence,

$\rm\implies \:sinA \: + \: cosA \: > \: 1$

So, option (B) is correct.

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

Answer:

Given :

• If A is an acute angle in a right ABC, right angled at B.

To Find :-

• What is the value of Sin A + Cos A.

Solution :-

$\bigstar$ If A is an acute angle in a right ABC, right angled at B.

As we know that :

$\longrightarrow \sf\boxed{\bold{\pink{Sin\: A =\: \dfrac{Opposite\: Side}{Hypotenuse}}}}$

Given :

• Opposite Side = BC
• Hypotenuse = AC

According to the question by using the formula we get,

$\implies \sf Sin\: A =\: \dfrac{BC}{AC}$

$\implies \sf\bold{\purple{Sin\: A =\: \dfrac{BC}{AC}}}$

Again, we know that :

$\longrightarrow \sf\boxed{\bold{\pink{Cos\: A =\: \dfrac{Adjacent\: Side}{Hypotenuse}}}}$

Given :

• Adjacent Side = AB
• Hypotenuse = AC

According to the question by using the formula we get,

$\implies \sf Cos\: A =\: \dfrac{AB}{AC}$

$\implies \sf\bold{\purple{Cos\: A =\: \dfrac{AB}{AC}}}$

Now, we have to find the value of Sin A + Cos A :

Given :

$\bullet \: \: \bf Sin\: A =\: \dfrac{BC}{AC}$

$\bullet \: \: \bf Cos\: A =\: \dfrac{AB}{AC}$

So, by putting the values we get,

$\dashrightarrow \sf\bold{\green{Sin\: A + Cos\: A}}$

$\dashrightarrow \sf \dfrac{BC}{AC} + \dfrac{AB}{AC}$

$\dashrightarrow \sf \dfrac{BC + AB}{AC}$

As we know that :

$\leadsto$ The sum of two sides of a triangle is greater than the third side.

So,

$\mapsto \sf\bold{\red{Sin\: A + Cos\: A >\: \dfrac{AC}{AC} =\: 1}}$

$\therefore$ The value of Sin A + Cos A is greater than one .

Hence, the correct options is option no (B) greater than one .