Question

please solve my question correctly correctly correctly....don't spam.....explanation must...​

Answer 1

Hi !

Solution :-

From given :- tan⁴¢ + tan²¢. = 1

= tan⁴¢ + tan²¢ = tan²¢ ( tan²¢ + 1)

We know. , ( 1+tan²¢ ) = sec²¢

=> tan²¢ ( sec²¢) = 1

=> (sec²¢ - 1) sec²¢ = 1

=> sec⁴¢ - sec²¢ = 1

=> 1/cos⁴¢ - 1/cos²¢ = 1

=> 1- cos²¢ / cos⁴¢ = 1

=> sin²¢ /cos⁴¢ = 1

=> sin²¢ = cos⁴¢ .....______ __ (1)

Now , we have to prove .

Cos⁴ ¢ + cos²¢ = 1

cos⁴¢ = 1-cos²¢

cos⁴¢ = sin²¢ .

Since , from equation (1) we have already proved from given that , sin²¢ = cos⁴¢

°•° cos⁴¢ + cos²¢ =1

from LHS

= sin²¢ + cos²¢ = 1 ,{From equation (1/

= 1 proved

LHS = RHS

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Hope it's helpful

Answer 2

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies {tan}^{4} \theta + {tan}^{2} \theta = 1 \\ \\ \red{\underline \bold{To \: Prove: }} \\ \tt: \implies {cos}^{4} \theta + {cos}^{2} \theta =1$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {tan}^{4} \theta + {tan}^{2} \theta = 1 \\ \\ \tt: \implies {tan}^{2} \theta( {tan}^{2} \theta + 1) = 1 \\ \\ \tt \circ \: tan^{2} \theta = {sec}^{2} \theta-1$

$\tt: \implies ({sec}^{2} \theta - 1)sec^{2} \theta = 1 \\ \\ \tt: \implies {sec}^{4} \theta - {sec}^{2} \theta = 1 \\ \\ \tt: \implies \frac{1}{ {cos}^{4} \theta} - \frac{1}{ {cos}^{2} \theta} = 1 \\ \\ \tt: \implies \frac{1 - {cos}^{2} \theta }{ {cos}^{4} \theta } = 1 \\ \\ \tt \circ {sin}^{2} \theta = 1 - {cos}^{2} \theta \\ \\ \tt: \implies \frac{ {sin}^{2} \theta }{ {cos}^{4} \theta} = 1 \\ \\ \tt: \implies {sin}^{2} \theta = {cos}^{4} \theta - - - - - (1) \\ \\ \bold{To \: Prove : } \\ \tt: \implies {cos}^{4} + {cos}^{2} \theta = 1 \\ \\ \bold{Proof : } \\ \tt: \implies {cos}^{4} \theta + {cos}^{2} \theta \\ \\ \text{Putting \: value \: of \: {cos}}^{2} \theta \\ \tt: \implies {sin}^{2} \theta + {cos}^{2} \theta \\ \\ \tt \circ \: {sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \\ \tt: \implies 1 \\\\ \: \: \: \: \: \huge\green{\bold{Proved}}$