Prove that (sin A + cosec A)² +(cos A + sec A)² = 7 + tan²A + co²A.
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Answers
Step-by-step explanation:
(sinA+cscA) ^2+(cosA+secA) ^2
=sin ^2 A+csc ^2 A+2sinAcscA+cos ^2
A+sec ^2 A+2cosAsecA .......As[a²+b²+2ab=(a+b)²]
=sin ^2 A+csc ^2 A+2sinA× 1 / sinA +cos ^2 A+sec ^2 A+2cosA 1 / cosA
........... since secA= 1 / cosA and cscA= 1 / sinA
=sin ^2 A+csc ^2 A+2+cos ^2 A+sec ^2 A+2
=(sin ^2 A+cos ^2 A)+csc ^2 A+sec ^2 A+4
=1+1+cot ^2 A+1+tan A+4 ...........
since csc ^2 A=1+cot ^2 A , and sec 2 A=1+tan ^2 A
=7+tan ^2 A+cot ^2 A
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Hence proved.
Answer:
( sin A + cosec A )² + ( cos A + sec A )² = 7 + tan² A + cot² A
Step-by-step-explanation:
We have to prove that, ( sin A + cosec A )² + ( cos A + sec A )² = 7 + tan² A + cot² A.
Now, the given trigonometric equation is
( sin A + cosec A )² + ( cos A + sec A )² = 7 + tan² A + cot² A.
LHS = ( sin A + cosec A )² + ( cos A + sec A )²
⇒ LHS = ( sin A )² + 2 * sin A * cosec A + ( cosec A )² + ( cos A )² + 2 * cos A * sec A + ( sec A )² - - - [ ∵ ( a + b )² = a² + 2ab + b² ]
⇒ LHS = sin² A + 2 sin A cosec A + cosec² A + cos² A + 2 cos A sec A + sec² A - - - [ ∵ ( sin A )² = sin² A ]
⇒ LHS = sin² A + cos² A + cosec² A + sec² A + 2 sin A * 1 / sin A + 2 cos A * 1 / cos A - - [ ∵ cosec A = 1 / sin A & sec A = 1 / cos A ]
⇒ LHS = 1 + cosec² A + sec² A + 2 * sin A ÷ sin A + 2 * cos A ÷ cos A - - - [ ∵ sin² A + cos² A = 1 ]
⇒ LHS = 1 + cosec² A + sec² A + 2 * 1 + 2 * 1
⇒ LHS = 1 + cosec² A + sec² A + 2 + 2
⇒ LHS = 1 + 2 + 2 + cosec² A + sec² A
⇒ LHS = 3 + 2 + cosec² A + sec² A
⇒ LHS = 5 + cosec² A + sec² A
⇒ LHS = 5 + ( 1 + cot² A ) + sec² A - - - [ ∵ cosec² A = 1 + cot² A ]
⇒ LHS = 5 + 1 + cot² A + sec² A
⇒ LHS = 6 + cot² A + ( 1 + tan² A ) - - - [ ∵ sec² A = 1 + tan² A ]
⇒ LHS = 6 + cot² A + 1 + tan² A
⇒ LHS = 6 + 1 + cot² A + tan² A
⇒ LHS = 7 + cot² A + tan² A
⇒ LHS = 7 + tan² A + cot² A
RHS = 7 + tan² A + cot² A
∴ LHS = RHS