Question

sec (65degree +thitha)-cosec(25degree -thitha)​

Answer 1

Given Question :-

Find the value of

$\tt \: sec(65\degree \: + \theta \:) - cosec(25\degree \: - \theta \:)$

$\large\underline{\bold{Solution-}}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$(1). \: \boxed{ \bf \: sec(90\degree \: - \: x) = cosecx}$

Given that

$\tt \: sec(65\degree \: + \theta \:) - cosec(25\degree \: - \theta \:) - - (1)$

$\bf \: Let \: 65\degree \: + \theta \: = x$

and

$\bf \: Let \: 25\degree \: - \theta \: = y$

So,

$\rm :\implies\:x + y = 65\degree \: + \theta \: + 25\degree \: - \theta \:$

$\rm :\implies\:x + y = 90\degree \:$

$\rm :\implies\:y = 90\degree \: - x$

$\rm :\implies\:secy = sec(90\degree \: - x)$

$\rm :\implies\:secy = cosecx$

$\bf\implies \:cosecx \: - \: secy \: = \: 0$

Put values of x and y, we get

$\bf :\implies\:cosec(65\degree \: + \theta \:) - sec(25\degree \: - \theta \:) = 0$

Additional Information :-

$(1). \: \boxed{ \bf \: sin(90\degree \: - \theta \:) = \: cos\theta \:}$

$(2). \: \boxed{ \bf \: cos(90\degree \: - \theta \:) = \: sin\theta \:}$

$(3). \: \boxed{ \bf \: tan(90\degree \: - \theta \:) = \: cot\theta \:}$

$(4). \: \boxed{ \bf \: cot(90\degree \: - \theta \:) = \: tan\theta \:}$

$(5). \: \boxed{ \bf \: cosec(90\degree \: - \theta \:) = \: sec\theta \:}$

Answer 2

Step-by-step explanation:

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