Question

If sinA+cosA=3/2 then the value of sum of all trigonometric ratio with angle A

Answer 1

Answer:

11/2

Step-by-step explanation:

Hi,

Given sin A + cos A = 3/2

Now squaring on both sides we get

sin²A + cos²A +2 sin A cos A = 9/4

⇒ 1 + 2 sin A cos A = 9/4

⇒2 sin A cos A = 5/4

⇒ sin A cos A = 5/8

Now,

Consider sin A + cos A + tan A + cot A + cosec A + sec A

= (sin A + cos A) + (sin A/cos A + cos A/sin A) + (1/sin A + 1/ cos A)

=(sin A + cos A) + (sin²A + cos²A)/sin A cos A + (sin A + cos A)/sin A cos A

= 3/2 + 8/5 + 8/5*3/2

= 3/2 + 8/5 + 12/5

= 11/2

Hope, it helped !

Answer 2

Answer :

$sin \: A + cos \: A + tan \: A+ cot \: A \\+ sec \: A + cosec \: A =\frac{11}{2}$

Solution:

Given
$sin \: A + cos \: A = \frac{3}{2} \: \: \: eq1$

To find:

$sin \: A + cos \: A + tan \: A+ cot \: A + sec \: A + cosec \: A$

if we can convert the complete statement in the form of sin A and cos A,then only it can be solved

As we know that tan A,cot A,sec A and Cosec A can be written in the form of Sin A and Cos A

So,

$sin \: A + cos \: A + \frac{sin \: A}{cos \:A } + \frac{cos \: A}{sin \: A} + \frac{1}{cos \: A} + \frac{1}{sin \: A} \\ \\ = sin \: A + cos \: A + \frac{ {sin}^{2}A + {cos}^{2}A + sin \: A + cos \: A}{sin \: A \: cos \: A} \\ \\ since \\ \\ {sin}^{2} A + {cos}^{2} A = 1 \\ \\ = (sin \: A + cos \: A) + (\frac{1 + sin \: A + cos \: A }{sin \: A \: cos \: A})$
Here sin A+ cos A is given,to find sin A.cos A,
square both sides of the eq1

$( {sin \: A + cos \: A)}^{2} = \frac{3}{4} \\ \\ {sin}^{2} A + {cos}^{2} A + 2sin \: A\: cos \: A = \frac{9}{4} \\ \\ 2sin \: A \: cos \: A = \frac{9}{4} - 1 \\ \\ sin \: A \: cos \: A = \frac{5}{8}$
put this value and value from eq 1 in the expression

$= [sin \: A + cos \: A] + (\frac{1 + [sin \: A + cos \: A ]}{sin \: A \: cos \: A}) \\ \\ = \frac{3}{2} + \frac{ 1 + \frac{3}{2} }{ \frac{5}{8} } \\ \\ = \frac{3}{2} + \frac{ \frac{5}{2} }{ \frac{5}{8} } \\ \\ = \frac{3}{2} + 4 \\ \\ = \frac{11}{2}$

Hope it helps you.