Question

sec theta - 10 theta = a so prove sec theta = 1/2 [a+1/a] and tan theta = 1/2 [ 1/a - a ]​

Answer 1

Answer:

ʜᴇʏᴀ!!!

Step-by-step explanation:

sec^2 theta - tan^2 theta = 1

tan^2 theta= sec^2 theta-1

sec theta - √ sec^2 theta -1 = a

(sec theta -a)^2 = sec^2 theta -1

sec^2 theta + a.a - 2a sec the ta = sec^2 theta -1

a.a - 2a sec theta = -1

sec theta= a.a +1)/2a

= 1/2 ( a+1/a)

Answer 2

From the principles of trigonometry we know that :-

sec²θ - tan²θ = 1

We know that by expansions : a² - b² = ( a + b )( a - b )

So we can write the trigonometric identity as :-

( secθ + tanθ )( secθ - tanθ ) = 1

Given in the question secθ - tanθ = a

So basically we will have :-

a ( secθ + tanθ ) = 1

⇒ secθ + tanθ = 1/a

We have two equations :

Adding them :

secθ - tanθ = a ..............( 1 )

secθ + tanθ = 1/a ............( 2 )

------------------------

2 secθ = a + 1/a

⇒ secθ = 1/2 [ a + 1/a ]

Subtracting both :-

secθ - tanθ = a ..............( 1 )

secθ + tanθ = 1/a ............( 2 )

-------------------------

- 2 tanθ = a - 1/a

⇒ 2 tanθ = 1/a - a

⇒ tanθ = 1/2 [ 1/a - a ]

Hence they are proved !