if tan=15/8 and cos negative then what is the value of [sin(-a)-cosa]/[tan(-a)+sec(-a)]
Then tan is +, and the cos is -, so you must be in quadrant III
tan a = 15/8 = y/x
so y = -15 , x = -8, and r = 17
[sin(-a)-cosa]/[tan(-a)+sec(-a)]
= (-sina - cosa)/(-tana + seca)
= (15/17 + 8/17)/(-15/8 - 17/8)
= (23/17)(-8/32)
= -23/68
Thank you sir for solving this sum....... if we follow the trigonometric way then it will be very easy
This isnt the correct way u wont get the full marks for this sum if follow the above way... It should be soved by using the trigonometric formula
Well, since you mentioned "cos negative," I'm guessing you're in a bit of a negative mood. Don't worry, I've got your back! Let's break down the expression and have some fun with it:
First, let's simplify the numerator: sin(-a)-cos(a).
Now, if "a" is negative, then we can find some humor in the situation. It's like being in a room with negative people – they always seem to bring us down! So, let's call this expression "The Dejected Duo."
Next, for the denominator: tan(-a) + sec(-a).
Hmm, dealing with negative angles can be quite tricky, just like navigating through a maze of negativity. Hence, let's refer to this expression as "The Pessimistic Pair."
Adding a little humor to math, let's evaluate the value of our new expressions:
The Dejected Duo (numerator) = sin(-a) - cos(a)
Well, without diving into the math, we can say that the value of The Dejected Duo depends on how much disappointment and confusion is in the air! So, it's hard to assign an actual value to it, but I assure you, it's not having the best day.
Next, The Pessimistic Pair (denominator) = tan(-a) + sec(-a)
Considering the negative angle, things aren't looking very positive for us. Therefore, let's just say that this expression is like trying to find happiness in a rainstorm – it's unlikely to happen!
So, ultimately, the value of the entire expression [sin(-a) - cos(a)] / [tan(-a) + sec(-a)] is just a reflection of the negative vibes floating around. Remember, though, it's always better to laugh and find joy even in the face of negativity. Stay positive!
To find the value of the expression \(\frac{{\sin(-a)-\cos(a)}}{{\tan(-a)+\sec(-a)}}\), we'll begin by determining the values of \(\sin(-a)\), \(\cos(a)\), \(\tan(-a)\), and \(\sec(-a)\). Let's break the solution down step by step:
Step 1: Given that \(\tan = \frac{15}{8}\), we can use the Pythagorean identity \(\tan^2(a) + 1 = \sec^2(a)\) to calculate the value of \(\sec^2(a)\).
\(\tan^2(a) + 1 = \sec^2(a)\)
\(\left(\frac{15}{8}\right)^2 + 1 = \sec^2(a)\)
\(\frac{225}{64} + 1 = \sec^2(a)\)
\(\frac{289}{64} = \sec^2(a)\)
Step 2: Since we are given that \(\cos(a)\) is negative, we know that \(a\) is in either the second or third quadrant. In these quadrants, \(\cos(a)\) is negative. Therefore, \(\cos(a) = -\sqrt{1 - \sin^2(a)}\).
Step 3: We'll calculate \(\sin(a)\) using the value of \(\tan(a)\) we were given.
\(\sin(a) = \frac{{\tan(a)}}{{\sqrt{1 + \tan^2(a)}}}\)
\(\sin(a) = \frac{{\frac{15}{8}}}{{\sqrt{1 + \left(\frac{15}{8}\right)^2}}}\)
Step 4: Now, let's use the values we calculated in steps 2 and 3 to find the value of \(\sin(-a)\), \(\cos(a)\), \(\tan(-a)\), and \(\sec(-a)\).
\(\sin(-a)\) is the negative value of \(\sin(a)\).
\(\cos(-a)\) is equal to \(\cos(a)\) (since cosine is an even function).
\(\tan(-a)\) is the negative value of \(\tan(a)\).
\(\sec(-a) = \frac{1}{\cos(-a)}\).
Step 5: Substitute the calculated values into the expression \(\frac{{\sin(-a)-\cos(a)}}{{\tan(-a)+\sec(-a)}}\) and solve.
\(\frac{{\sin(-a)-\cos(a)}}{{\tan(-a)+\sec(-a)}} = \frac{{-\sin(a)-\cos(a)}}{{-\tan(a)+\sec(a)}}\)