Question

If ‘a’ is an acute angle such that 1 + cot a – cosec a = root 3- 1, then the value of 1 + tan a + sec a is?

Answer 1

Answer:1+√3

Step-by-step explanation

We substitute a values with 30, 45,60 (since a>90)

when a=

1+cot30-cosec30

1+root3-2

root3-1

Now we found that a=30

then 1+tan30+sec30

1+1/root3+2/root3

root3+1+2/root3

LCM as root3

root3+3/root3

factorization

root3+3/root3*root3/root3=3+root3/

3 wiil cancel and the answer is 1+root3

Answer 2

The value of ($1+ tana-seca$) is, $\frac{2}{\sqrt{3}-1 }$.

• Sine and cosine are trigonometric functions of an angle in mathematics. In the context of a right triangle, the sine and cosine of an acute angle are defined as the ratio of the length of the side directly opposite the angle to the length of the longest side of the triangle (the hypotenuse), and the neighbouring leg's length to the hypotenuse, respectively.
• The definitions of sine and cosine can be expanded more broadly to include any real value in terms of the lengths of certain line segments in a unit circle. The sine and cosine can be extended to arbitrary positive and negative values as well as complex numbers according to more recent definitions that represent them as infinite series or as the solutions to certain differential equations.
• Now, tangent can be defined as the ratio of these sine and cosine of any angle.

Here, we are given that,

$1+ cota-coseca = \sqrt{3} -1$

Now, multiplying ($1+ cota-coseca$) with ($1+ tana-seca$), we get,

$(\frac{sina+cosa-1}{sina} )(\frac{sina+cosa+1}{cosa} )\\=\frac{1+2sinacosa-1}{sinacosa} \\=\frac{2sinacosa}{sinacosa} \\=2$

Now, since the value of $1+ cota-coseca = \sqrt{3} -1$, we get the value of

($1+ tana-seca$) as, $\frac{2}{\sqrt{3}-1 }$.

Hence, the value of ($1+ tana-seca$) is, $\frac{2}{\sqrt{3}-1 }$.

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