Question

If SecA+TanA=x , find SinA and CotA​

Answer 1

Question

If sec A + tan A = x  

find sin A and cot A.

Solution

we have,

→  sec A + tan A = x   ...( Name it as eqn 1 )

squaring both sides

→  sec²A + tan²A + 2 sec A tan A = x²

using identity : sec² A= 1 + tan²A

→  1 + tan²A + tan²A + 2 sec A tan A = x²

→  1 + 2 tan²A + 2 sec A tan A = x²

→  2 tan²A + 2 sec A tan A = x² - 1

taking 2 tan A common in LHS

→ 2 tan A ( tanA + secA ) = x² - 1

using: sec A + tan A = x (given)

→  2 tan A  ·  x  = x² - 1

→  2 tan A = ( x² - 1 ) / x

→ tan A = ( x² - 1 ) / ( 2x )

so, → cot A = 2x / (x² - 1 )

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Let us now find the value of sin A

as we know

→  sec² A - tan² A = 1

→  ( sec A+ tan A)(sec A - tan A) = 1

→  x  ( sec A - tan A ) = 1

→ ( sec A - tan A ) = 1 / x  .... eqn (2)

adding eqn (1) and (2)

sec A + tan A + sec A - tan A = x + (1/x)

2 sec A = ( x² + 1 ) / x

→ sec A = ( x² + 1 ) / 2x

so,

→ cos A = 2x / (x²+1)

Now,

as we know

tan A = sin A / cos A

→ sin A = tan A · cos A

→ sin A = (x²-1)/2x  ·  2x / (x²+1)

→ sin A = (x²-1)/(x²+1)

Answer 2

$\sin A= \frac{p}{h} = \frac{ {x}^{2} - 1}{{x}^{2} + 1} \\ \\ \cot A = \frac{2x}{{x}^{2}-1}$

Step-by-step explanation:

Given:

sec A + tan A = x ---(1)

To find:

sinA and cotA

Solution:

As we know,

sec² A - tan² A = 1

or, (sec A + tan A)(sec A - tan A) = 1

or, x(sec A - tan A) = 1 [from ---(1)]

or, sec A - tan A = 1/x ----(2)

Subtracting (2) from (1) ,

$( \sec A + \tan A) - ( \sec A - \tan A ) = x - \frac{1}{x} \\ \\ </p><p> \sec A + \tan A - \sec A + \tan A = \frac{ {x}^{2} - 1 }{x} \\ \\ 2\tan A = \frac{ {x}^{2} - 1 }{x} \\ \\ \tan A = \frac{ {x}^{2} - 1 }{2x} \\ \\ \cot A = \frac {2x}{ {x}^{2} - 1}$

Now,

$\tan A = \frac{ {x}^{2} - 1 }{2x} = \frac{Perpendicular(p)}{base(b)} \\ \\ \therefore Hypotenuse (h) = \sqrt{ {p}^{2} + {b}^{2} } \\ \\ = \sqrt{( {x}^{2} - 1) {}^{2} + {(2x)}^{2} } \\ \\ = \sqrt{ {x}^{4} + 1 - 2 {x}^{2} + 4 {x}^{2} } \\ \\ = \sqrt{ {x}^{4} + 1 + 2 {x}^{2} } \\ \\ = \sqrt{( {x}^{2} + 1) {}^{2} } \\ \\ = {x}^{2} + 1 \\ \\ So, h = {x}^2 + 1$

Then,

$\sin A = \dfrac{p}{h} = \dfrac{ {x}^{2} - 1 }{ {x}^{2} + 1}$

Used Trigonometry Rules:

tanø . cotø = 1

sec²ø - tan²ø = 1

Used Algebra Rules:

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy$