Please answer my these questions,
there are more 3 Questions bit i can't attach them at once so plz first Answer this.......
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Answers
Question : -
( 1 + tan² A ) + ( 1 + 1/[ tan² A] ) = 1/( sin² A - sin⁴ A )
ANSWER
Given : -
( 1 + tan² A ) + ( 1 + 1/[ tan² A] ) = 1/( sin² A - sin⁴ A )
Required to prove : -
- LHS = RHS
Identities used : -
- sin² A + cos² A = 1
Proof : -
( 1 + tan² A ) + ( 1 + 1/[ tan² A] ) = 1/( sin² A - sin⁴ A )
Consider the LHS part
( 1 + tan² A) + ( 1 + 1/[ tan² A ] )
Since, we know that ;
- tan A = sin A/cos A
This implies;
( 1 + sin² A/cos² A ) + ( 1 + 1/[ sin² A/cos² A] )
( 1 + sin² A/cos² A ) + ( 1 + 1/1 x cos² A/sin² A )
( 1 + sin² A/cos² A ) + ( 1 + cos² A/sin² A )
By taking the LCM
( [cos² A + sin² A]/cos² A ) + ( [sin² A + cos² A]/sin² A )
From the identity;
- sin² A + cos² A = 1
1/cos² A + 1/sin² A
By taking LCM
( sin² A + cos² A)/( sin² A cos² A )
From the identity ;
- sin² A + cos² A = 1
- cos² A = 1 - sin² A
1/( sin² A [ 1 - sin² A ])
1/( sin² A - sin⁴ A )
Consider the RHS part ;
1/( sin² A - sin⁴ A )
>>> LHS = RHS <<<
Hence Proved !
Question : -
Note :-
Here, @ = theta
( sin @ - cos @ + 1)/(sin @ + cos @ - 1) = 1/(sec @ - tan @)
ANSWER
Given : -
( sin @ - cos @ + 1)/(sin @ + cos @ - 1) = 1/(sec @ - tan @)
Required to prove : -
- LHS = RHS
Identity used : -
- sec² @ - tan² @ = 1
Proof : -
( sin @ - cos @ + 1)/(sin @ + cos @ - 1) = 1/(sec @ - tan @)
Consider the LHS part;
( sin @ - cos @ + 1)/(sin @ + cos @ - 1)
Divide the numerator & denominator with cos @
( [sin @ - cos @ + 1]/[cos @] )/( [sin @ + cos @ - 1]/[cos @] )
( [sin @]/[cos @] - [cos@]/[cos @] + 1/cos @ )/( [sin @]/[cos @] + [cos @]/[cos @] - 1/cos @)
Since, we know that ;
- tan @ = sin @/cos @
- sec @ = 1/cos @
( tan @ - 1 + sec @ )/( tan @ + 1 - sec @ )
( tan @ + sec @ - 1 )/( tan @ - sec @ + 1 )
From the identity;
- sec² @ - tan² @ = 1
( tan @ + sec @ - 1 )/( tan @ - sec @ + sec² @ - tan² @)
( tan @ + sec @ - 1 )/( sec @ - tan @ - [ {sec @}² - {tan @}² ])
( tan @ + sec @ - 1)/( sec @ - tan @ - [ sec @ + tan @ ] [ sec @ - tan @ ] )
Taking ( sec @ - tan @ ) common in denominator
( tan @ + sec @ - 1 )/( sec @ - tan @ [ - 1 + sec @ + tan @ ] )
( tan @ + sec @ - 1 )/( sec @ - tan @ [ tan @ + sec @ - 1 ] )
tan @ + sec @ - 2 get's cancelled in both numerator and denominator
1/(sec @ - tan @)
Consider the RHS part;
1/( sec @ - tan @ )
>>> LHS = RHS <<<
Hence Proved !
Question : -
( cot A - cos A )/( cot A + cos A ) = ( cosec A - 1 )/( cosec A + 1 )
ANSWER
Given : -
( cot A - cos A )/( cot A + cos A ) = ( cosec A - 1 )/( cosec A + 1 )
Required to prove : -
- LHS = RHS
Proof : -
( cot A - cos A )/( cot A + cos A ) = ( cosec A - 1 )/( cosec A + 1 )
Consider the LHS part ;
( cot A - cos A )/( cot A + cos A )
Since, we know that;
- cot A = cos A/sin A
( [cos A]/[sin A] - cos A )/( [cos A]/[sin A] + cos A )
By taking LCM in both numerator and denominator
( [cos A - cos A sin A ]\[ cos A ] )/( [cos A + cos A sin A ]/[cos A] )
cos A gets cancelled;
( cos A - cos A sin A )/( cos A + cos A sin A )
Taking cos A common on both sides;
( cos A [ 1 - sin A ]/cos A [ 1 + sin A ] )
cos A gets cancelled in both numerator and denominator
( 1 - sin A )/( 1 + sin A )
Dividing the numerator & denominator with sin A
( 1/sin A - [sin A ]/[ sin A ] )/( 1/sin A + [sin A ]/[sin A ] )
( cosec A - 1 )/(cosec A + 1 )
Consider the RHS part;
( cosec A - 1 )/(cosec A + 1 )
>>> LHS = RHS <<<
Hence Proved !
Answer: