Question

(sec²ø-1) (cosex²ø-1)​

Answer 1

Answer:

(sec²ø-1) (cosec²ø-1)

= (tan²ø)(cot²ø)

Because, sec²ø = 1 + tan²ø

cosec²ø = 1 + cot²ø

= (tan²ø)(1/tan²ø)

Because, cot²ø = 1/tan²ø

= 1

Thus, the correct answer is 1.

Answer 2

Answer:

$\displaystyle{\boxed{\red{\sf\:(\:\sec^2\:\theta\:-\:1\:)\:(\:cosec^2\:\theta\:-\:1\:)\:=\:1\:}}}$

Step-by-step-explanation:

We have given a trigonometric expression.

We have to find the value of that expression.

The given trigonometric expression is

$\displaystyle{\sf\:(\:\sec^2\:\theta\:-\:1\:)\:(\:cosec^2\:\theta\:-\:1\:)}$

$\displaystyle{\implies\sf\:\sec^2\:\theta\:(\:cosec^2\:\theta\:-\:1\:)\:-\:1\:(\:cosec^2\:\theta\:-\:1\:)}$

$\displaystyle{\implies\sf\:\sec^2\:\theta\:\times\:cosec^2\:\theta\:-\:\sec^2\:\theta\:-\:cosec^2\:\theta\:+\:1}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\sec\:\theta\:=\:\dfrac{1}{\cos\:\theta}\:}}}$

$\displaystyle{\implies\sf\:\left(\:\dfrac{1}{\cos\:\theta}\:\right)^2\:\times\:cosec^2\:\theta\:-\:\left(\:\dfrac{1}{\cos\:\theta}\:\right)^2\:-\:cosec^2\:\theta\:+\:1}$

$\displaystyle{\implies\sf\:\dfrac{1}{\cos^2\:\theta}\:\times\:cosec^2\:\theta\:-\:\dfrac{1}{\cos^2\:\theta}\:-\:cosec^2\:\theta\:+\:1}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:cosec\:\theta\:=\:\dfrac{1}{\sin\:\theta}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{1}{\cos^2\:\theta}\:\times\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)^2\:-\:\dfrac{1}{\cos^2\:\theta}\:-\:\left(\:\dfrac{1}{\sin\:\theta}\:\right)^2\:+\:1}$

$\displaystyle{\implies\sf\:\dfrac{1}{\cos^2\:\theta}\:\times\:\dfrac{1}{\sin^2\:\theta}\:-\:\dfrac{1}{\cos^2\:\theta}\:-\:\dfrac{1}{\sin^2\:\theta}\:+\:1}$

$\displaystyle{\implies\sf\:\dfrac{1}{\cos^2\:\theta\:.\:\sin^2\:\theta}\:-\:\dfrac{\sin^2\:\theta\:-\:\cos^2\:\theta}{\cos^2\:\theta\:.\:\sin^2\:\theta}\:+\:1}$

$\displaystyle{\implies\sf\:\dfrac{1\:-\:(\:\sin^2\:\theta\:-\:\cos^2\:\theta\:)}{\cos^2\:\theta\:.\:\sin^2\:\theta}\:+\:1}$

$\displaystyle{\implies\sf\:\dfrac{1\:-\:\sin^2\:\theta\:+\:\cos^2\:\theta\:}{\cos^2\:\theta\:.\:\sin^2\:\theta}\:+\:1}$

We know that,

$\displaystyle{\boxed{\green{\sf\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1\:}}}$

$\displaystyle{\implies\sf\:\dfrac{1\:-\:1}{\cos^2\:\theta\:.\:\sin^2\:\theta}\:+\:1}$

$\displaystyle{\implies\sf\:\dfrac{0}{\cos^2\:\theta\:.\:\sin^2\:\theta}\:+\:1}$

$\displaystyle{\implies\sf\:0\:+\:1}$

$\displaystyle{\implies\sf\:1}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:(\:\sec^2\:\theta\:-\:1\:)\:(\:cosec^2\:\theta\:-\:1\:)\:=\:1\:}}}}$