Solve : ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )
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Solve : ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )
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Given :-
To Do :-
Solution :-
❍ To calculate the given expression we must know that ::
Finding the solution :-
⇝ ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )
⇝ sec²θ( 1 - sin²θ )( 1 + sin²θ )
⇝ sec²θ( 1² - sin²θ )
⇝ sec²θ × ( 1 - sin²θ )
⇝ sec²θ × cos²θ
⇝ 1/cos²θ × cos²θ
⇝ 1
More to know :-
Note : To see clearly, view the answer on brainly.in/question/41920659
Answer:
Given :-
( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )
To Do :-
Solve it.
Solution :-
❍ To calculate the given expression we must know that ::
1 + tan²θ = sec²θ
( a + b )( a - b ) = a² - b²
1 - sin²θ = cos²θ
Finding the solution :-
⇝ ( 1 + tan²θ )( 1 - sin²θ )( 1 + sin²θ )
According to the information we have, we can write the first bracket as sec²θ
⇝ sec²θ( 1 - sin²θ )( 1 + sin²θ )
According to the property of ( a + b )( a - b ) = a² - b²
⇝ sec²θ( 1² - sin²θ )
⇝ sec²θ × ( 1 - sin²θ )
According to the information we have, we can write the bracket as cos²θ
⇝ sec²θ × cos²θ
⇝ 1/cos²θ × cos²θ
⇝ 1