Math, asked by aditya63439, 1 day ago

pls any matha genius help me with these sums pls beg u​

Answers

Answered by BrainlyPARCHO
2

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

L. H. S =

(cosecA - sinA)×(secA - cosA)

= (1/sinA - sinA)×(1/cosA - cosA)

= (1-sin²A)/sinA × (1-cos²A)/cosA

= cos²A/sinA × sin²A/cosA

= cos²Asin²A/cosAsinA

= cosAsinA

Now, R. H. S =

1/(tanA + cotA)

= 1/(sinA/cosA + cosA/sinA)

= 1/[(sin²A + cos²A)/cosAsinA]

= 1/[(1/cosAsinA)]

= cosAsinA

Since L. H. S. = R. H. S.

Therefore the given equation is proof.

Answered by BrainTeach
1

\mathbb\blue{HELLO\:\:\:\:DEAR\:\:\:\:USER}

\Large{\underline{\underline{\sf{\red{Req}\pink{uir}\green{ed}\:\purple{Ans}\blue{wer}\:\orange{:-}}}}}

L. H. S =

(cosecA - sinA)×(secA - cosA)

= (1/sinA - sinA)×(1/cosA - cosA)

= (1-sin²A)/sinA × (1-cos²A)/cosA

= cos²A/sinA × sin²A/cosA

= cos²Asin²A/cosAsinA

= cosAsinA

Now, R. H. S =

1/(tanA + cotA)

= 1/(sinA/cosA + cosA/sinA)

= 1/[(sin²A + cos²A)/cosAsinA]

= 1/[(1/cosAsinA)]

= cosAsinA

Since L. H. S. = R. H. S.

Therefore the given equation is proof.

\Large{\underline{\underline{\sf{\red{B}\pink{e}\green{\:}\:\purple{Brai}\blue{nly}\:\orange{!}}}}}

Similar questions